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Table of contents :
Cover
Title
Copyrights
Kishore Vaigyanik Protsahan Yojana
CONTENTS
QUESTION PAPER 2019
QUESTION PAPER 2018
QUESTION PAPER 2017KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA (Nov 19)
QUESTION PAPER 2017KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA (Nov 05)
QUESTION PAPER 2016KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2015KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2014KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2013KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2012KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2011KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2010KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
QUESTION PAPER 2009KISHORE VAIGYANIK PROTSAHAN YOJANAStream : SA
PRACTICE SET 1
PRACTICE SET 2
PRACTICE SET 3
PRACTICE SET 4
PRACTICE SET 5
Kishore Vaigyanik Protsahan Yojana
11 Years’
SOLVED PAPERS
2019  2009
Stream SA
Included 5 Practice Sets
Kishore Vaigyanik Protsahan Yojana
11 Years’
SOLVED PAPERS
2019  2009
Stream SA Authors Lakshman Prasad (Mathematics) Deepak Paliwal, Mansi Garg (Physics) Neha Minglani Sachdeva (Chemistry) Sanubia Saleem (Biology)
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Kishore Vaigyanik Protsahan Yojana
ABOUT THE EXAM KVPY i.e. Kishore Vaigyanik Protsahan Yojana is a National Level Fellowship (scholarship) Program in Basic Science (Physics, Chemistry, Mathematics & Biology) upto PrePhd Level, run by Department of Science & Technology, Government of India and Conducted by IISC (Indian Institute of Science) Bangalore, Karnataka Annually. It Was Started in 1999 to Encourage Basic Sciences Students to take up Research Career in Sciences. The Objective of the Exam is to Encourage Talented Students for Research Career in Sciences.
ELIGIBILITY CRITERIA KVPY scholarships are given only to Indian Nationals to study in India. There are three streams in KVPY; SA, SB & SX. Eligibility criteria for different streams is discussed below; Ÿ
For SA Class 11 Students who passed class 10 with minimum 75% (65% for SC/ST/PWD) marks in Mathematics & Science. The fellowship of students selected in SA will be activated only if they pursue undergraduate courses in Basic Sciences (B.Sc./B.S./B.Stat./B.Math/Integrated M.Sc. or M.S.) and have secured a minimum of 60% (50% for SC/ST/PWD) marks in science subjects in class 12th.
Ÿ
For SX Class 12 Students aspiring to pursue undergraduate program (B.Sc. etc ) with basic sciences (Physics, Chemistry, Mathematics & Biology) who passed class 10 with minimum 75% (65% for SC/ST/PWD) marks in Mathematics & Science. The fellowship of students selected in SX will be activated only if they pursue undergraduate courses in Basic Sciences (B.Sc./B.S./B.Stat./B.Math/Integrated M.Sc. or M.S.) and have secured a minimum of 60% (50% for SC/ST/PWD) marks in science subjects in class 12th.
Ÿ
For SB B.Sc. Ist year Students who passed class 12 with 60% marks in Maths & Sciences (PCMB) & class 10 with minimum 75% marks in Mathematics & Science. In order to activate fellowship, in the first year of undergraduate course they should secure minimum 60% (50% for SC/ST/PWD) marks.
Those students who are intending or pursuing undergraduate program under distance education scheme or correspondence course of any university are not eligible.
SYLLABUS OF KVPY There is no prescribed syllabus for KVPY aptitude test, it aims to assess the understanding & analytical ability of the students than his/her factual knowledge. However questions are framed from syllabus upto class 10/12/Ist Year of Undergraduate Courses in basic sciences, as applicable. There are two Questions Papers in KVPY; one for stream SA & Other for SB/SX (Question Paper is same for SB & SX).
QUESTION PAPERS PATTERN There are two Questions Papers in KVPY; one for stream SA & Other for SX/SB (Question Paper is same for SB & SX). Ÿ
Question Paper for SA Stream caries 80 Questions for 100 marks. There are Two Parts in the Question Paper; Part I has 15 Questions of 1 mark each for Mathematics, Physics, Chemistry & Biology while Part II has 5 Questions of 2 marks each for Mathematics, Physics, Chemistry & Biology.
Ÿ
Question Paper for SB/SX Stream caries 120 Questions for 160 marks. There are Two Parts in the Question Paper; Part I has 20 Questions of 1 mark each for Mathematics, Physics, Chemistry & Biology while Part II has 10 Questions of 2 marks each for Mathematics, Physics, Chemistry & Biology.
MODE OF EXAM KVPY is conducted in Online Mode in English & Hindi Medium.
TIME OF EXAM Ÿ
Normally the notification or advertisement for KVPY appear in National Newspapers on May 11 (Technology Day) and Second Sunday of July every year.
Ÿ
Generally the exam is conducted in the month of November.
SELECTION PROCESS After scrutiny of application forms on the basis of eligibility criteria for various streams all eligible students are called for Aptitude Test conducted in English & Hindi Medium at different centers across the country. On the basis of performance in aptitude test shortlisted students are called for an interview, which is the final stage of selection procedure.
FELLOWSHIPS The selected students are eligible to receive KVPY fellowship after class 12th/Ist Year of Undergraduate course only if they pursue Undergraduate Courses in Basic Science, upto PrePhD or 5 Years whichever is earlier. Details of fellowships are listed below; Basic Science
Annual Contingency Grant
Monthly Fellowship
SA/SX/SB during Ist to IIIrd year of B.Sc./B.S./BB.Stat./B.Math/ Integrated M.Sc or M.S.
Rs. 5000
Rs. 20000
SA/SX/SB during M.Sc. / IVth to Vth years of Integrated M.Sc /M.S./ M.Math/ M.Stat.
Rs. 7000
Rs. 28000
CONTINUATION / RENEWAL OF FELLOWSHIP Ÿ The fellow should continue to study basic science and should also maintain a minimum level of academic performance as Ist division or 60% (50% for SC/ST/PWD) marks in aggregate. Also the fellow has to pass all the subjects prescribed for that particular year. Ÿ In each year marks are to be certified by the Dean or Head of the Institution. Ÿ The fellowship will be discontinued if above marks are not obtained. However if fellow passed all the subjects & obtain marks more than 60% (50% for SC/ST/PWD) in subsequent year, the fellowship can be renewed only for that year onwards. Ÿ If KVPY fellow opts out of the basic science at any stage then monthly fellowship and contingency grant will be forfeited from him.
KVPY Timeline 2020 IMPORTANT DATES Opening of Application Portal Last Date of Submission of Online Application KYPY Aptitude Test APPLICATION FEE For General Category For SC/ST/ Persons with Disabilities For more details visit:www.kvpy.iisc.ernet.in
: :
2nd Week of July 2020 1st Week of September 2020
:
1st Week of November 2020
: :
Rs. 1000/Rs. 500/
CONTENTS KVPY SA QUESTION PAPERS (20192009) QUESTION PAPER
QUESTION PAPER
QUESTION PAPER
2019
2018
2017 (19 Nov)
Pg. No. 116
Pg. No. 115
Pg. No. 1631
QUESTION PAPER
QUESTION PAPER
QUESTION PAPER
2017 (05 Nov)
2016
2015
Pg. No. 3247
Pg. No. 4862
Pg. No. 6376
QUESTION PAPER
QUESTION PAPER
QUESTION PAPER
2014
2013
2012
Pg. No. 7792
Pg. No. 93107
Pg. No. 108122
QUESTION PAPER
QUESTION PAPER
QUESTION PAPER
2011
2010
2009
Pg. No. 123137
Pg. No. 138150
Pg. No. 151164
KVPY PRACTICE SETS (15)
167232
KVPY Question Paper 2019 Stream : SA
1
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2019 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS
Then, which of the following statements are true?
1. Let ABC be an equilateral triangle with side length a. Let R and r denote the radii of the circumcircle and the incircle of triangle ABC respectively. Then, as a R function of a, the ratio r (a) strictly increases (b) strictly decreases (c) remains constant (d) strictly increases for a < 1and strictly decrease for a > 1
2. Let b be an nonzero real number. Suppose the 1 quadratic equation 2x + bx + = 0 has two distinct b real roots. Then 2
(a) b +
1 5 > b 2
(c) b2 − 3b > − 2
1 5 < b 2 1 (d) b2 + 2 < 4 b
(b) b +
3. Let p(x) = x + ax + b have two distinct real roots, 2
where a , b are real number. Define g(x) = p(x3 ) for all real number x.
I. g has exactly two distinct real roots. II. g can have more than two distinct real roots. III. There exists a real number α such that g(x) ≥ α for all real x. (a) Only I (c) Only II
(b) Both I and III (d) Both II and III
4. Let a n , n ≥ 1, be an arithmetic progression with first term 2 and common difference 4. Let M n be the 10
average of the first n terms. Then the sum
∑ M n is
n =1
(a) 110 (c) 770
(b) 335 (d) 1100
5. In a triangle ABC, ∠BAC = 90°; AD is the altitude from A on to BC. Draw DE perpendicular to AC and DF perpendicular to AB. Suppose AB = 15 and BC = 25. Then the length of EF is (a) 12 (c) 5 3
(b) 10 (d) 5 5
KVPY Question Paper 2019 Stream : SA
2 6. The sides a , b, c of a triangle satisfy the relations c2 = 2ab and a 2 + c2 = 3b2. Then the measure of ∠BAC, in degrees, is (a) 30 (c) 60
(b) 45 (d) 90
7. Let N be the least positive integer such that whenever a nonzero digit c is written after the last digit of N, the resulting number is divisible by c. The sum of the digits of N is (a) 9 (c) 27
(b) 18 (d) 36
15. The number of three digit numbers abc such that the
8. Let x1, x2,…, x11 be 11 distinct positive integers. If we replace the largest of these integers by the median of the other 10 integers, then (a) the median remains the same (b) the mean increases (c) the median decreases (d) the mean remains the same
(a) 0 (b) 1 (c) more than one but finitely many (d) infinitely many
10. A twodigit number ab is called almost prime if one obtains a twodigit prime number by changing at most one of its digits a and b. (For example, 18 is an almost prime number because 13 is a prime number). Then the number of almost prime twodigit numbers is (c) 87
(d) 90
11. Let P be an interior point of a convex quadrilateral ABCD and K , L , M , N be the midpoints of AB, BC, CD, DA respectively. If Area (PKAN ) = 25, Area (PLBK ) = 36, and Area (PMDN ) = 41 then Area (PLCM ) is (a) 20
(b) 29
(c) 52
(a) 9 (c) 36
(b) 18 (d) 54
16. Various optical processes are involved in the
P(1) = 2, P(2) = 4, P(3) = 6, P(4) = 8 is
(b) 75
arithmetic mean of b and c and the square of their geometric mean are equal is
PHYSICS
9. The number of cubic polynomials P (x) satisfying
(a) 56
(b) given any positive real number α, we can choose C and Area (C ) is less than α T as above such that ratio Area (T ) Area (C ) (c) give any C and T as above, the ratio is Area (T ) independent of C and T (d) there exist real numbers a and b such that for any circle C and triangle T as above, we must have Area (C ) a< 0 be a real number, C denote a circle with circumference l and T denote a triangle with perimeter l. Then (a) given any positive real number α, we can choose C and Area (C ) is greater than α T as above such that ratio Area (T )
0
0
(a) 0.26 m/s (c) 0.45 m/s
0.5
1 Time (s)
1.5
(b) 0.33 m/s (d) 0.21 m/s
2
KVPY Question Paper 2019 Stream : SA 19. A student in a town in India, where the price per unit (1 unit = 1 kWh) of electricity is ` 5.00, purchases a 1 kVA UPS (uninterrupted power supply) battery. A day before the exam, 10 friends arrive to the student’s home with their laptops and all connect their laptops to the UPS. Assume that each laptop has a constant power requirement of 90 W. Consider the following statements.
I. All the 10 laptops can be powered by the UPS, if connected directly. II. All the 10 laptops can be powered, if connected using an extension box with a 3 A fuse. III. If all the 10 friends use the laptop for 5 h, then the cost of the consumed electricity is about ` 22.50. Select the correct option with the true statements. (a) I only (c) I and III only
(b) I and II only (d) II and III only
20. Frosted glass is widely used for translucent windows. The region, where a transparent adhesive tape is stuck over the frosted glass becomes transparent. The most reasonable explanation for this is (a) diffusion of adhesive glue into glass (b) chemical reaction at adhesive tapeglass interface (c) refractive index of adhesive glue is close to that of glass (d) adhesive tape is more transparent than glass
21. Consider two equivalent, triangular hollow prisms A and B made of thin glass plates and arranged with negligible spacing as shown in the figure. A beam of white light is incident on prism A from the left. Given that, the refractive index of water is inversely related to temperature, the beam to the right of prism B would not appear white, if
te Whi
t ligh
A
B
(a) both prisms are filled with hot water (70°C) (b) both prisms are filled with cold water (7°C) (c) both prisms are empty (d) prism A is filled with hot water (70°C) and prism B with cold water (7°C)
3 hemisphere at a height h from the horizontal surface, then the speed of the particle is (a) (b) (c) (d)
(2g (R − h )) (2g (R + h )) 2gR 2gh
24. The nuclear radius is given by R = r0 A1/ 3 , where r0 is constant and A is the atomic mass number. Then, the nuclear mass density of U238 is (a) twice that of Sn119 (b) thrice that of Sn119 (c) same as that of Sn119 (d) half that of Sn119
25. The electrostatic energy of a nucleus of charge Ze is kZ 2e2 , where k is a constant and R is the R nuclear radius. The nucleus divides into two Ze and equal radii. The daughter nuclei of charges 2 change in electrostatic energy in the process when they are far apart is
equal to
0.375kZ 2e2 R kZ 2e2 (c) R
0125 . kZ 2e2 R 0.5kZ 2e2 (d) R (b)
(a)
26. Two masses M1 and M 2 carry positive charges Q1 and Q2, respectively. They are dropped to the floor in a laboratory set up from the same height, where there is a constant electric field vertically upwards. M1 hits the floor before M 2. Then, (a) Q1 > Q2 (c) M1Q1 > M2Q2
(b) Q1 < Q2 (d) M1Q2 > M2Q1
27. Which one of the following schematic graphs best represents the variation of pV (in Joules) versus T (in Kelvin) of one mole of an ideal gas? (The dotted line represents pV = T )
(a) pV (J)
(b) pV (J)
T (K)
22. A ball is moving uniformly in a circular path of
T (K)
radius 1 m with a time period of 1.5 s. If the ball is suddenly stopped at t = 83 . s, the magnitude of the displacement of the ball with respect to its position at t = 0 s is closest to (a) 1 m (c) 3 m
(b) 33 m (d) 2 m
(c) pV (J)
(d) pV (J)
23. A particle slides from the top of a smooth hemispherical surface of radius R which is fixed on a horizontal surface. If it separates from the
T (K)
T (K)
KVPY Question Paper 2019 Stream : SA
4 28. Mumbai needs 14 . × 1012 L of water annually. Its
34. IUPAC name of the following compound
effective surface area is 600 km 2 and it receives an average rainfall of 2.4 m annually. If 10% of this rain water is conserved, it will meet approximately (a) 1% of Mumbai’s water needs (b) 10% of Mumbai’s water needs (c) 50% of Mumbai’s water needs (d) 100% of Mumbai’s water needs
HO
is
29. A mass M moving with a certain speed V collides elastically with another stationary mass m. After the collision, the masses M and m move with speeds V ′ and v, respectively. All motion is in one dimension. Then, (a) V = V ′ + v (V + v) (c) V ′ = 2
(b) V ′ = V + v (d) v = V + V ′
face PQ of an isosceles prism PQR with apex angle ∠Q = 120°. The refractive indices of the material of the prism for the above rays 1, 2, 3 and 4 are 1.85, 1.95, 2.05 and 2.15 respectively and the surrounding medium is air. Then, the rays emerging from the face QR are (b) 1 and 2 only (d) 1, 2, 3 and 4
CHEMISTRY 31. The hybridisations of N, C and O shown in the following compound R C
N
(a) sp , sp, sp (c) sp 2, sp, sp
from the reaction of steam with (a) methane (c) carbon monoxide
(b) coke (d) carbon dioxide
caused by (a) CaCl 2 (c) Ca(HCO3 )2
(b) CaSO4 (d) CaCO3
37. The most polarisable ion among the following is (a) F−
(b) I−
(c) Na +
(d) Cl −
38. For a multielectron atom, the highest energy level among the following is 1 2 1 (b) n = 4, l = 2, m = 0, s = + 2 1 (c) n = 4, l = 1, m = 0, s = + 2 1 (d) n = 5, l = 1, m = 0, s = + 2 (a) n = 5, l = 0, m = 0, s = +
(a) As2O3 2
2
2
35. In watergas shift reaction, hydrogen gas is produced
39. The oxide, which is neither acidic nor basic is
O
respectively, are 2
(a) 1hydroxycyclohex4en3one (b) 1hydroxycyclohex3en5one (c) 3hydroxycyclohex5en1one (d) 5hydroxycyclohex2en1one
36. Treatment with lime can remove hardness of water
30. Four ray 1, 2, 3 and 4 are incident normally on the
(a) 4 only (c) 3 and 4 only
O
(b) sp , sp , sp (d) sp, sp, sp 2
2
(c) N2O
(d) Na 2O
flame test is (a) Mg
32. The following compounds
(b) Sb4 O10
40. The element whose salts cannot be detected by (b) Na
(c) Cu
(d) Sr
41. The plot of concentration of a reactant vs time for a
are (a) geometrical isomers (c) optical isomers
(b) positional isomers (d) functional group isomers
33. The major product of the following reaction Br Br
1. Excess alc. KOH
O
2. NaNH2
Ph
is H H H
(b) Ph
isolated chamber,
Br H
(c) Ph
H (d) Ph
Br
(a) 0 (b) 1 (c) 2 (d) not possible to determine from this plot
42. During the free expansion of an ideal gas in an Br
Br
Time
The order of this reaction with respect to the reactant is
+ 3. H3O
(a) Ph
Concentration
chemical reaction is shown below.
H
(a) internal energy remains constant (b) internal energy decreases (c) work done on the system is negative (d) temperature increases
KVPY Question Paper 2019 Stream : SA 43. The number of moles of water present in a spherical
5 51. The mode of speciation mediated by geographical
water droplet of radius 1.0 cm is,
isolation is referred as
[Given : Density of water in the droplet = 10 . g cm −3 ]
(a) adaptive radiation (b) allopatric speciation (c) parapatric speciation (d) sympatric speciation
π (a) 18 (c) 24π
2π (b) 27 2π (d) 9
44. Among the following, the correct statement about cathode ray discharge tube is (a) the electrical discharge can only be observed at high pressure and at low voltage. (b) in the absence of external electrical or magnetic field, cathode rays travel in straight lines. (c) the characteristics of cathode rays depend upon the material of electrodes. (d) the characteristics of cathode rays depend upon the gas present in the cathode ray tube.
45. For a spontaneous process, (a) enthalpy change of the system must be negative (b) entropy change of the system must be positive (c) entropy change of the surrounding must be positive (d) entropy change of the system plus surrounding must be positive
BIOLOGY 46. Which one of the following is a CORRECT statement about primates’ evolution? (a) Chimpanzees and gorillas evolved from macaques (b) Humans and chimpanzees evolved from gorillas (c) Human, chimpanzees and gorillas evolve from a common ancestor (d) Humans and gorillas evolved from chimpanzees
47. The crypts of Lieberkuhn are found in which one of the following parts of the human digestive tract? (a) Oesophagus (b) Small intestine (c) Stomach (d) Rectum
(a) Glucose to pyruvate (b) Glucose to CO 2 and ethanol (c) Glucose to lactate (d) Glucose to CO 2 and H2 O
53. Where are the proximal and distal convoluted tubules located within the human body? (a) Adrenal cortex (b) Adrenal medulla (c) Renal cortex (d) Renal medulla
54. In a diploid organism, when the locus X is inactivated, transcription of the locus Y is triggered. Based on this observation, which one of the following statements is CORRECT? (a) X is dominant over Y (b) X is epistatic to Y (c) Y is dominant over X (d) Y is epistatic to X
55. Which one of the following sequences represents the CORRECT taxonomical hierarchy? (a) Species, genus, family, order (b) Order, genus, family, species (c) Species, order, genus, family (d) Species, genus, order, family
56. Which one of the following organs is NOT a site for the production of white blood cells? (a) Bone marrow (c) Liver
(b) Kidney (d) Spleen
involved in guttation?
(a) lipids and carbohydrates only (b) lipids and proteins only (c) lipids, proteins and carbohydrates (d) proteins and carbohydrates only
49. Microscopic examination of a blood smear reveals an abnormal increase in the number of granular cells with multiple nuclear lobes. Which one of the following cell types has increased in number? (b) Monocytes (d) Thrombocytes
50. Which one of the following genetic phenomena is represented by the blood group AB? (a) Codominance (c) Overdominance
requires oxygen?
57. Which one of the following anatomical structures is
48. Removal of the pancreas impairs the breakdown of
(a) Lymphocytes (c) Neutrophils
52. Which one of the following metabolic conversion
(b) Dominance (d) Semidominance
(a) Cuticle (c) Lenticels
(b) Hydathodes (d) Stomata
58. Which one of the following parts of the eye is affected in cataract? (a) Cornea (c) Retina
(b) Conjunctiva (d) Lens
59. Which one of the following organisms is a bryophyte? (a) Liverwort (c) Chlamydomonas
(b) Volvox (d) Fern
60. During oogenesis in mammals, the second meiotic division occurs (a) before fertilisation (b) after implantation (c) before ovulation (d) after fertilisation
KVPY Question Paper 2019 Stream : SA
6
PARTII (2 Marks Questions) MATHEMATICS
67. In an hourglass approximately 100 grains of sand
61. Let a , b, c, d be distinct real numbers such that a , b are roots of x2 − 5cx − 6d = 0, and c, d are roots of x2 − 5ax − 6b = 0. Then b + d is (a) 180
(b) 162
(c) 144
(d) 126
62. Let S = {1, 2, 3, … , 100}. Suppose b and c are chosen at random from the set S. The probability that 4x2 + bx + c has equal roots is (a) 0.001
(b) 0.004
(c) 0.007
(d) 0.01
63. Let N be the set of positive integers. For all n ∈ N , let fn = (n + 1)1/ 3 − n1/ 3 and 1 A = n ∈ N : fn + 1 < < fn 2/3 3(n + 1) Then,
fall per second (starting from rest); and it takes 2 s for each sand particle to reach the bottom of the hourglass. If the average mass of each sand particle is 0.2 g, then the average force exerted by the falling sand on the bottom of the hourglass is close to (a) 0.4 N (c) 1.2 N
68. A student uses the resistance of a known resistor
(1 Ω ) to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be 1 × 103 Ω in circuit (a) and 0999 . Ω in circuit (b). From these measurements, the resistance (in Ω) of the voltmeter and ammeter are found to be close to v
primes p1, p2, p3 , p4 such that p = p1 + p2 = p3 − p4. The number of special primes is
65. Let ABC be a triangle in which AB = BC. Let X be a point on AB such that AX : XB = AB : AX. If AC = AX, then the measure of ∠ABC equals (c) 54°
(d) 72°
(a) 2
−2
(a) 10 and 10 (c) 10−2 and 102
(b)
(b) 10 and 10−3 (d) 10−2 and 103 3
Assume that the balloon is spherical in shape with diameter of 11.7 m and the mass of the balloon and the payload (without the hot air inside) is 210 kg. Temperature and pressure of outside air are 27 °C and 1 atm = 105 N/m 2, respectively. Molar mass of dry air is 30 g. The temperature of the hot air inside is close to [The gas constant, R = 831 . JK −1mol−1] (a) 27 °C (c) 105 °C
(b) 52 °C (d) 171 °C
70. A healthy adult of height 1.7 m has an average blood
PHYSICS 66. A waterproof laser pointer of length 10 cm placed in a water tank rotates about a horizontal axis passing through its centre of mass in a vertical plane as shown in the figure. The time period of rotation is 60 s. Assuming the water to be still and no reflections from the surface of the tank, the duration for which the light beam escapes the tank in one time period is close to (Take, refractive index of water = 133 . )
40cm
30cm
Laser
10cm
(a) 8.13 s (c) 16.67 s
1=W
69. A hot air balloon with a payload rises in the air.
(a) 0 (b) 1 (c) more than one but finite (d) infinite
(b) 36°
1=W
v
64. A prime number p is called special if there exist
A
v
(a) A = N (b) A is a finite set (c) the complement of A in N is nonempty, but finite (d) A and its complement in N are both infinite
(a) 18°
(b) 0.8 N (d) 1.6 N
30cm
(b) 14.05 s (d) 23.86 s
pressure (BP) of 100 mm of Hg. The heart is typically at a height of 1.3 m from the foot. Take, the density of blood to be 103 kg/m3 and note that 100 mm of Hg is equivalent to 13.3 kPa (kilo pascals). The ratio of BP in the foot region to that in the head region is close to (a) one (c) three
(b) two (d) four
CHEMISTRY 71. PbO2 is obtained from (a) the reaction of PbO with HCl (b) thermal decomposition of Pb(NO3 )2 at 200°C (c) the reaction of Pb3 O4 with HNO3 (d) the reaction of Pb with air at room temperature
KVPY Question Paper 2019 Stream : SA
(a) ‘‘b’’ increases and ‘‘a’’ decreases at constant temperature (b) ‘‘b’’ decreases and ‘‘a’’ increases at constant temperature (c) temperature increases at constant ‘‘a’’ and ‘‘b’’ values (d) ‘‘b’’ increases at constant ‘‘a’’ and temperature
73. The correct statements among the following. i. E 2s (H) > E 2s (Li) < E 2s (Na) > E 2s (K). ii. The maximum number of electrons in the shell with principal quantum number n is equal to 2n 2 . iii. Extra stability of halffilled subshell is due to smaller exchange energy.
genotype for male and XX for female. What will be the genotype of the embryos and endosperm nuclei after double fertilisation? (a) 50% ovules would have XXX endosperm and YY embryo, while the other 50% would have XXY endosperm and XX embryo (b) 100% ovules would have XXX endosperm and XY embryo (c) 100% ovules would have XXY endosperm and XX embryo (d) 50% ovules would have XXX endosperm and XX embryo, while the other 50% would have XXY endosperm and XY embryo
78. Solid and dotted lines represent the activities of pepsin and salivary amylase enzymes of the digestive tract, respectively. Which one of the following graphs best represents their activity vs pH?
(a)
iv. Only two electrons, irrespective of their spin, may exist in the same orbital are.
1
(b) ii and iii (d) i and iv
74. An organic compound contains 46.78% of a halogen X. When 2.00 g of this compound is heated with fuming HNO3 in the presence of AgNO3 , 2.21 g AgX was formed. The halogen X is [Given : atomic weight of Ag = 108, F = 19, Cl = 355 ., Br = 80, I = 127] (a) F
(b) Cl
(c) Br
(d) I
75. An organic compound X with molecular formula C6H10, when treated with HBr, forms a gemdibromide. The compound X upon warming with HgSO4 and dil. H2SO4, produces a ketone, which gives a positive iodoform test. The compound X is
(c)
5 pH
10
Activity
(a) i and ii (c) iii and iv
(b)
(d)
1
5 pH
Activity
[Given : ‘‘a’’ and ‘‘b’’ are standard parameters for van der Waals’ gas]
77. Papaya is a dioecious species with XY sexual
1
5 pH
10
1
5 pH
10
Activity
pV compressibility factor Z = at a fixed volume RT will certainly decrease, if
Activity
72. For one mole of a van der Waals’ gas, the
7
10
79. If the gene pool of the locus X in the human genome is 4, then what would be the highest possible number of genotypes in a large population? (a) 6 (c) 10
(b) 8 (d) 16
80. Match the plant hormones in Column I with their primary function in Column II.
(a)
(b)
(c)
(d)
C
BIOLOGY 76. A cell weighing 1 mg grows to double its initial mass before dividing into two daughter cells of equal mass. Assuming no death, at the end of 100 divisions what will be the ratio of the mass of the entire population of these cells to that of the mass of the earth? Assume that mass of the earth is 1024 kg and 210 is approximately equal to 1000. (a) 10−28 (c) 1
(b) 10−3 (d) 103
Column I
Column II
P.
Abscisic acid
i.
Promotes disease resistance
Q.
Ethylene
ii.
Maintains seed dormancy
R.
Cytokinin
iii.
Promotes seed germination
S.
Gibberellin
iv.
Promotes fruit ripening
v.
Inhibits leaf senescence
Choose the correct combination (a) P–iii, Q–iv, R–i, S–ii (b) P–ii, Q–iv, R–v, S–iii (c) P–v, Q–iii, R–ii, S–i (d) P–iv, Q–ii, R–iii, S–v
KVPY Question Paper 2019 Stream : SA
8
Answers PARTI 1
(c)
2
(c)
3
(b)
4
(a)
5
(a)
6
(b)
7
(a)
8
(c)
9
(a)
10
(d)
11
(c)
12
(c)
13
(d)
14
(a)
15
(b)
16
(a)
17
(a)
18
(b)
19
(c)
20
(c)
21
(d)
22
(d)
23
(a)
24
(c)
25
(a)
26
(d)
27
(a)
28
(b)
29
(d)
30
(c)
31
(a)
32
(d)
33
(a)
34
(d)
35
(d)
36
(c)
37
(b)
38
(b,d)
39
(c)
40
(a)
41
(a)
42
(a)
43
(b)
44
(b)
45
(d)
46
(c)
47
(b)
48
(c)
49
(c)
50
(a)
51
(b)
52
(d)
53
(c)
54
(d)
55
(a)
56
(b)
57
(b)
58
(d)
59
(a)
60
(d)
PARTII 61
(c)
62
(a)
63
(a)
64
(b)
65
(b)
66
(c)
67
(a)
68
(b)
69
(c)
70
(c)
71
(c)
72
(b)
73
(a)
74
(c)
75
(d)
76
(c)
77
(d)
78
(a)
79
(c)
80
(b)
Solutions 1. (c) For an equilateral triangle ABC having side length a. If R and r are radii of the circumcircle and the incircle of triangle ABC respectively, then a a 2 a R = sec 30° = = 2 3 3 2 A
R B
30° a/ 2
r M
60° a/ 2
C
a a 1 a tan 30° = × = 2 2 3 2 3 a R ∴ = 3 = 2, which is independent of a a r 2 3 and it is constant. and r =
2. (c) Given quadratic equation 1 2x + bx + = 0, has two distinct real b roots, so D>0 1 2 b − 4(2) > 0 ⇒ b 2
b3 − 8 8 > 0⇒ >0 b b (b − 2)(b2 + 2b + 4) ⇒ >0 b ⇒ b ∈ (−∞ , 0) ∪ (2, ∞ ) For option (c), b2 − 3b > − 2 ⇒ b2 − 3b + 2 > 0 ⇒ (b − 2)(b − 1) > 0 b ∈ (−∞ , 1) ∪ (2, ∞ ) mean if b ∈ (−∞ , 0) ∪ (2, ∞ ) then b2 − 3b > − 2 ⇒
b2 −
…(i)
3. (b) Let the given quadratic polynomial p (x) = x2 + ax + b has two distinct real roots α and β, then p (x) = x2 + ax + b = (x − α )(x − β ) and since g (x) = p (x3 ) = (x3 − α )(x3 − β ) let α = α31 and β = β31 then g (x) = (x3 − α31 )(x3 − β31 ) = (x − α1 )(x − β1 )(x2 + α1 x + α12 ) (x2 + β1 x + β12 ) Q the discriminants of quadratic equations x2 + α1 x + α12 and x2 + β1 x + β12 are negative. ∴ g (x) has exactly two distinct real roots and since g (x) = x6 + ax3 + b is an even degree polynomial, so there exists a real number ‘α’ such that g (x) ≥ α for all real x.
4. (a) The sum of first n ,n ≥ 1terms of arithmetic progression with first term 2 and common difference 4, is n Sn = [4 + (n − 1)4] = 2n 2 2 So, the average of the first n terms S Mn = n = 2n n Now,
10
10
n =1
n =1
∑ Mn = 2 ∑ n 10 × 11 = 2 × = 110 2
5. (a) It is given that in triangle ABC, ∠BAC = 90°, AD is the altitude from A on to BC. B D
F
A
C
E
Since, AB = 15 and BC = 25 ∴
AC = =
BC 2 − AB 2 =
625 − 225
400 = 20 1 (BC )(AD ) 2 1 = (AB )(AC ) 2
Now, since area of ∆ABC =
KVPY Question Paper 2019 Stream : SA 1 1 ⇒ (BC )(AD ) = × 15 × 20 2 2 ⇒ 25 × AD = 300 ⇒ AD = 12 Q AEDF is a rectangle, then EF = AD = 12
6. (b) It is given that the sides of triangle, a , b and c satisfy the following relations …(i) c2 = 2ab 2 2 2 and …(ii) a + c = 3b From Eqs. (i) and (ii), we get A
Ö2a a
C
⇒ ⇒ ⇒ ∴
a
B
a 2 + 2ab = 3b2 a + 2ab + b2 = 4b2 (a + b)2 = (2b)2 = (b + b)2 a = b, so c = 2a ∠A = ∠B = 45° 2
7. (a) As N be the least positive integer and when a nonzero digit C is written after the last digit of N, the resulting number is divisible by C. So, 10N + C is divisible by C ∴10N must be divisible by C. Now, the least integer (N ) which is divisible by digit ‘C’ i.e. (1 to 9) must be L.C.M of {1, 3, 4, 6, 7, 9}. = L.C.M of {4, 7, 9} = 252 = N and sum of digits of number ‘N’ is 2+ 5+ 2= 9
Now,
12. (c) Given equations
…(i) P (1) = a + b + c + d = 2 …(ii) P (2) = 8a + 4b + 2c + d = 4 P (3) = 27a + 9b + 3c + d = 6 …(iii) P (4) = 64a + 16b + 4c + d = 8 …(iv) From Eqs. (i) and (ii), we get …(v) 7a + 3b + c = 2 From Eqs. (ii) and (iii), we get …(vi) 19a + 5b + c = 2 From Eqs. (iii) and (iv), we get …(vii) 37a + 7b + c = 2 Now, from Eqs. (v) and (vi), we get …(viii) 12a + 2b = 0 and from Eqs. (vi) and (vii), we get …(ix) 18a + 2b = 0 From Eqs. (viii) and (ix), we get a = 0 and b = 0, c = 2 and d = 0. So, P (x) = 2x ∴no cubic polynomial is possible.
…(i) 6x + 4 y + z = 200, and x + y + z = 100 …(ii) By Eqs. (i) and (ii), we get 5x + 3 y = 100 For nonnegative integer solutions, when x = 2, then y = 30 x = 5, then y = 25 x = 8, then y = 20 x = 11, then y = 15 x = 14, then y = 10 x = 17, then y = 5 and x = 20, then y = 0 In every case z = 100 − (x + y) > 0 So, total number of nonnegative integral solutions are 7.
10. (d) Since in the group of first 10 two digit number 1019, has atleast 1 prime number similarly in other groups of 10 two digits numbers 2029, 3039, 4049, 5059, 6069, 7079, 8089 and 9099 have almost 1 prime numbers. So, the number of almost prime twodigit number is 90. 11. (c) Let a convex quadrilateral ABCD and K , L, M, N be the midpoint of AB, BC, CD, DA respectively.
9. (a) Let the equation of a cubic polynomial P (x) = ax3 + bx2 + cx + d
A
K B
x
x
w
y L
y
C
w
P z
8. (c) Let the given 11 distinct positive integers are in increasing order x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 , x10 , x11 , so x11 is largest of these integers and the median is x6 . Now, median of first 10 numbers is x6 + x6 = m (Let). 2 Now, we have to replace largest number x11 by m and then increasing order will be x1 , x2 , x3 , x4 , x5 , m, x6 , x7 , x8 , x9 , x10 x + x6 < x6 Q m < x6 as x5 < 5 2 So, median decreases.
9
N
z M
D
Now, as area ∆AKP = area ∆BKP = x (let) Similarly ∆BLP = ∆CLP = y ∆CPM = ∆DPM = z and ∆DNP = ∆ANP = w It is given that Area (PKAN ) = x + w = 25 area (PLBK ) = x + y = 36 and area (PMDN ) = z + w = 41 So area (PLCM ) = y + z = ( x + y ) + ( z + w) − ( x + w) = area (PLBK ) + area (PMDN ) − area (PKAN ) = 36 + 41 − 25 = 77 − 25 = 52
13. (d) It is given that, N 2 = 165 = 3 × 5 × 11and N1 = 255 + 1 As we know that, if n is odd integer then xn + yn is divisible by x + y. So, N1 = 255 + 155 is divisible by 2 + 1 = 3 and N1 = 255 + 155 = (25 )11 + (15 )11 = (32)11 + (1)11 is divisible by 32 + 1 = 33 ∴the HCF of N1 and N 2 is 33.
14. (a) It is given that circumference of circle C is l and the perimeter of triangle T is l. Now, let the radius of circle C is r, so l 2πr = l ⇒ r = 2π l2 ∴area of circle C is A1 = πr 2 = 4π Now, as we know that area of triangle will be maximum for given perimeter if it is an equilateral triangle, let the length of side of equilateral triangle is ‘a’, then l 3a = l ⇒ a = 3 and area of equilateral triangle is 3 2 A2 = a 4 l2 3 l2 So, A2 = = 4 9 12 3 l2 4π = 3 3 > 1 Q π l2 12 3 Since, as we took an equilateral triangle, which has maximum area. But we can take a triangle T such that the ratio area (C) is greater than any positive real area (T ) A1 = A2
number α.
KVPY Question Paper 2019 Stream : SA
10 15. (b) It is given that, the number of three digit number abc, such that b+ c …(i) = bc 2 the above relation is true if b = c = 0 And if neither b nor c is zero, then 1 1 + = 2, and b, c ∈{1, 2, 3, 4, 5, 6, 7, 8, 9} b c Then b = c = 1 and a ∈{1, 2, 3, 4, 5, 6, 7, 8, 9} So, total number of such three digit number are 2 × 9 = 18
16. (a) Formation of rainbow is shown below. Refraction
White light
Vio let
40°
t le
Total internal reflection
Refraction Re d
o Vi
Red
42°
Observer
So, processes involved in formation of rainbow in correct order are: refraction, total internal reflection, refraction. Hence, the correct order is given in option (a).
17. (a) Here, 10 divisions of vernier scall = 11main scale divisions So, 1 vernier scale division =
11 main 10
scale divisions Now, we use formula for least count, Least count = 1main scale division − 1 vernier scale division. ⇒ LC = 1MSD − 1VSD 11 = 1 − MSD 10
18. (b) Velocity = Slope of distance
− Time graph Last portion of given graph is a straight line which indicates that velocity is constant, i.e. terminal velocity is reached.
A B R
x2=0.4 te Whi
x1=0.3
1.6 1.9 t1 t2
(s)
From data of graph, Terminal velocity, x − x1 0.4 − 0.3 v= 2 = t2 − t1 1.9 − 1.6 0.1 = = 0.33 m/s 0.3
19. (c) Power requirement for 1 laptop, P1 = 90 W So, power requirement for 10 laptops, P = 10 × P1 = 10 × 90 = 900 W = 0.9 kW In 5h, electrical energy used by all laptops, E = P × t = 0.9 × 5 = 4.5 kWh Cost of electrical energy used is Cost = E × Unit cost = 4.5 × 5 = ` 22.50 So, statement III is correct. For laptop charger, input voltage is 220 V. So, current when all 10 laptops are connected through an extension, P 900 = ≈ 4.1A I= 220 V As, line current exceeds current rating of fuse, therefore 3A fuse cannot be used. So, statement II is incorrect.
20. (c) Frosted glass has a rough layer which causes irregular refraction and makes glass translucent. When a transparent tape which has refractive index close to that of glass is pasted over the rough surface of glass, the tape glue fills the roughness of glass. This makes glass surface more smooth and so refraction is more regular. This makes region of tape transparent.
21. (d) Prism B is inverted relative to prism A. So, dispersion of light caused by prism A and B is in opposite direction.
V
R V
If bending of light caused by B is less than or more than that of A, then out going beam of light is not white. So, when both prisms are filled with water at different temperatures, their refractive indices are different and the dispersion produced by A and B are not equal and opposite. Hence, with condition in (d) beam to right of prism B will be coloured.
22. (d) Time period of rotation of ball = 1.5 s So, in time interval of 7.5 s (= 1.5 × 5) s ball completes 5 revolutions. Also, ball covers onefourth of circular 1.5 path in time = 0.375 s. 4 So, in remaining 0.8 s (= 8.3 − 7.5s) ball is very near to other end of diameter as shown in the figure. Position of ball at t=0 and at t=7.5 s
Position of ball at t=7.5+0.375 =7.875s Position of ball at t=8.3s
nt me 3s ce t=8. a l p t Dis ball a of
Clearly, displacement of ball is nearly equals to diameter (= 2 m) of circular path.
23. (a) Condition given in question is shown below.
R–h R
Particle separates here
321
1 MSD 10 1 = − × 1mm 10 = −01 . mm So, magnitude of least count is 0.1 mm. =−
te Whi
D(m)
v h
Let v = speed of particle when it separates from hemisphere. As there is no friction, loss of potential energy appears in form of kinetic energy of particle.
KVPY Question Paper 2019 Stream : SA ∴ ⇒
1 mv2 2 v = 2 g (R − h )
mg (R − h ) =
11 2h anet
26. (d) Time of fall =
qE
24. (c) Given, nuclear radius is 1
+
R = r0 A 3 Here, atomic mass number of nucleus = A ∴Nuclear density d is given by Mass number d= Volume A A d= = ⇒ 1 4 3 4 πR π (r0 A 3 )3 3 3 A 3 ⇒ d= = 4 3 4 π r03 πr0 ⋅ A 3 As r0 = a constant, so nuclear density is a constant quantity. ∴ Nuclear mass density of U238 is same as that of Sn119 .
2 2
kZ e ...(i) U1 = R When this nucleus is divided into two equal nuclei of radius r, then as density of nuclear matter is a constant, we have initial density = final density M M = 2 4 3 4 3 πR πr 3 3 R3 R or r = 1 ...(ii) ⇒ r3 = 2 3 2 Now, final electrostatic energy is given by 2
Z 2k e2 2 kZ ′ 2e2 U2 = 2 × ⇒U 2 = r R 1 23 1 kZ 2e2 [from Eq. (ii)] = 2 ⋅ R 23 1 U 2 = 2 ⋅ U1 ⇒ 23 [from Eq. (i)] = 0.63U1 So, change in electrostatic energy in this process is ∆U = U1 − U 2 (QU1 > U 2 ) = U1 − 0.63U1 = (1 − 0.63) U1 = 0.375U1 kZ 2e2 [From Eq. (i)] = 0.375 R
1 1 > a1 a2
⇒
⇒ a2 > a1 When reciprocal is taken in equality sign is reversed, then. QE QE g − 1 > g− 2 M2 M1 Q1 E Q2E ⇒ − >− M1 M2 Q1 E Q2E < ⇒ M1 M2 Here, multiplication with − 1reverse sign of inequality. Q1 Q2 So, < M1 M2 M2Q1 < M1Q2 M1Q2 > M2Q1
or ⇒
27. (a) From gas equation, pV = nRT Here, n = 1mole So, pV = RT ...(i) Substituting the value of R in Eq. (i), we get pV = 8.3 T Clearly, slope of pV versus T line is 8.3, which is greater than one. Hence, following graph is correct. pV (J)
29. (d) Collision is elastic, so both linear momentum and kinetic energy are conserved. We have following situation, Before collision
M
After collision
1 2 3
of charge Ze is
Net acceleration of charged masses is qE anet = g − m As, M1 hits the floor before M2. 2h 2h > ⇒ a1 a2
1 2 3
25. (a) Electrostatic energy of a nucleus
E mg
28. (b) Surface area over which rain is received, A = 600 km 2 = 600 × (103 )2 m 2 = 6 × 108 m 2 Average rainfall, h = 2.4 m Volume of water received by rain, V = A × h = 6 × 108 × 2.4 m3 Water conserved = 10% of volume received by rain 10 = 6 × 108 × × 2.4 m3 = 1.44 × 108 m3 100 = 1.4 × 108 × 103 L = 1.4 × 1011 L Percentage of total water consumption received by rain is 1.4 × 1011 × 100 = 10% = 1.4 × 1012
M
V
V¢
m=0
m
v
According to figure, MV = MV ′ + mv ...(i) (linear momentum conservation) 1 1 1 ...(ii) MV 2 = MV ′2 + mv2 2 2 2 (kinetic energy conservation) From Eqs. (i) and (ii), we get ...(iii) M (V − V ′ ) = mv and M (V 2 − V ′2 ) = mv2 ...(iv) Dividing Eq. (iv) by Eq. (iii), we have M (V 2 − V ′2 ) mv2 = M (V − V ′ ) mv or V + V ′ = v
30. (c) Total internal, reflection occurs when n ≥
1 . sin ic
4 3
RT
Q
2
= pV R T e= V= p lo p
1 120°
3
S
S
4 30°
1 e= lop
P
T (K)
2 1
R
KVPY Question Paper 2019 Stream : SA
12 In given situation, angle of incidence of each of ray is 30° over face PR. So, i = 30° 1 1 ⇒ = =2 sin i sin 30° Hence, for total internal reflection at surface PR, n ≥ 2 . As refractive index for 3 and 4 is more than 2, only rays 1 and 2, pass from face PR while rays 3 and 4 pass through face QR (as shown in diagram).
31. (a) Hybridisation is determined from
N== C==O:
:
R
:
the steric number (number of atoms bonded to the central atom + the number of lone pairs). Number of hybrid orbitals must be equal to the steric number. From the Lewis structure.
(i) Steric number of Natom = 3 (2 bonded atoms + 1 lone pair), ∴Hybridisation = sp 2 (3 hybrid orbitals). (ii) Steric number of Catom = 2 (2 bonded atoms), ∴Hybridisation = sp (2 hybrid orbitals). (iii) Steric number of Oatom = 3 (1 bonded atom + 2 lone pair) ∴Hybridisation = sp 2 (3 hybrid orbitals).
Ph
Br
2.NaNH2 –HBr –2NH3
Ph—CºC Na
+ + 3.H3O
36. (c) Temporary hardness (caused by bicarbonates of calcium or magnesium) can be removed by using lime, Ca(OH)2. Ca(HCO3 )2 + Ca(OH)2 → 2CaCO3 + 2H2O
37. (b) Among anions with same charge, the one having greatest size has maximum polarisability. Thus, I− ion having most polarisability. 38. (b, d) Among the orbitals of a multielectron atom, the one with greatest value of n + l has the greatest energy. Between two orbitals with same value of n + l e.g. options (b) and (d), the one with greater value of n has greater energy.
41. (a)
Br Ph
CO2 + H2 In this reaction, hydrogen gas is produced from the reaction of steam with carbon dioxide.
40. (a) Of all the sblock elements, Mg and Be salts do not impart colour to flame.
33. (a) 1.Excess alcoholic KOH –HBr
CO + H2O →
neither acidic nor basic.
other one is an alkadiene. Since, they have two different functional groups, they are functional group isomers.
Br
FeO ⋅ Cr2 O3 (Catalyst)
39. (c) N2O is a neutral oxide, which is
32. (d) One isomer is an alkyne and the
Ph
35. (d) Watergas shift reaction is
Br
Ph—CºCH or PhºH
For a reaction X → Y , − d[X ] rate = ; [X ] = concentration of X. dt If reaction is nth order, rate ∝ [X ]n d[X ] From the graph, the slope is dt constant. ∴Rate is constant at any concentration. ∴ n=0
42. (a) In a free expansion, external
34. (d)
O 6 HO 5
1
2 3
4 Principal functional group is ketone. ∴C1 is carbonyl carbon atom. Locants for hydroxyl groups and double bonds are 5 and 2, which are preferred over 3 and 5, since the lower number at first difference (2 compared to 3) is preferred. Hence, the IUPAC name of given compound is 5hydroxycyclohex2en1one.
pressure ( pex ) = 0 ∴ W = − pex ⋅ ∆V = 0 and the system is isolated. Heat does not enter or leave, q = 0. ∆U = q + W = 0, whereU = internal energy.
43. (b) Number of moles, n=
mass (m) molar mass (M )
Given, radius = 1.0 cm, 4π cm3 ∴volume = 3 Given, density = 1.0 g cm − 3 ,
4π g. 3 (Atomic weight of water = 18 ) 4π 2π ∴ n= = 3 × 18 27 ∴Mass = volume × density =
44. (b) Cathode ray is observed only at low pressure and high voltage, which travel in straight line in the absence of electrical and magnetic fields. Characteristics of cathode rays are independent of the material of electrode or the gas present in the tube. 45. (d) For a spontaneous process in an isolated system, the change in entropy is positive, i.e, ∆S > O. However, if a system is not isolated, the entropy change of both the system and surroundings are to be taken into account because system and surroundings together constitute the isolated system thus, the total entropy change (∆Stotal ) is sum of the change in entropy of the system (∆Ssystem ) and the change in entropy of the surroundings (∆Ssurroundings ), i.e., ∆Stotal = ∆Ssystem + ∆SSurroundings for a spontenceus process, ∆Stotal must be positive, i.e., ∆Stotal is also termed as ∆Suniverse.
46. (c) The correct statement for primates’ evolution is that human, chimpanzees and gorillas share a common ancestor. From fossil records, primatologists came to know that human, chimpanzee and gorilla are evolved from a common ancient ancestor about 10 million years ago. Recent studies on gorilla genome confirmed that gorilla diverged from the common ancestor about 6 million years ago.
47. (b) The crypts of Lieberkuhn are found in small intestine. Crypts are invagination of the epithelium around the villi and lined largely with younger epithelial cell which are involved in secretion of mucus. 48. (c) Removal of pancreas impairs the breakdown of lipids, proteins and carbohydrates because pancreas produces insulin and other important enzyme like trypsin, chymotrypsin, amylase and lipase which helps in breakdown of macromolecules.
49. (c) Microscopic examination of blood smear reveals an abnormal increase in neutrophils. Neutrophils have a multilobed nucleus and granulated cytoplasm.
KVPY Question Paper 2019 Stream : SA Their number increases in blood in response of bacterial infection, acute inflammation and Eclampsia. Neutrophils are produced by hematopoiesis in the bone marrow and are active phagocytic cells. Lymphocytes are white blood cells which occurs in blood, lymph and lymphoid organs. Monocytes are mononuclear phagocytic cells. Platelets are known as thrombocytes and helps in blood clotting.
50. (a) Blood group AB represents codominance. In codominance a heterozygous individual expresses both alleles simultaneously with blending. No single allele is dominant over the other. Expression of both A and B alleles at same time results in AB type blood.
51. (b) Allopatric speciation is a genetic divergence permitted by geographical isolation. It is a speciation that occurs when population of the same species becomes isolated due to geographic barriers such as mountain ranges and water bodies. The population is reproductively isolated and each of the population accumulates different mutation and become diverge.
52. (d) Conversion of glucose to CO2 and H2O requires oxygen. In aerobic respiration glucose reacts with oxygen forming ATP, carbon dioxide and water are released as byproducts. C6 H12O6 + 6O2 → 6CO2 + 6H2O + ATP
53. (c) Proxima and distal convoluted tubules are located in renal cortex. Convoluted means the tubules one tightly coiled. Proximal convoluted tubules are associated with the reabsorption of filtered water, Na + , K+ . glucose, amino acid, Cl − , HCO3− , Ca 2 + , Mg 2 + and secretion of H+ , NH+4 , urea whereas distal convoluted tubules are associated with reabsorption of water, Na + , Cl − and Ca 2 + .
54. (d) When one gene masks or modifies the expression of another gene at distinct locus is known as epistasis. Gene that masks other or expresses itself is epistatic gene and gene that is masked is hypostatic gene. Here, X is inactivated by Y and triggers its own expression that means Y is epistatic to loci X because it masks the expression of X.
55. (a) The correct taxonomic hierarchy is species, genus, family, order. Taxonomy is the branch of biology that
deals with identification, nomenclature and classification. Carlous Linnaeus invented binomial nomenclature and developed a classification system known as taxonomic hierarchy. The various units of classification is kingdom, phylum, class, order, family, genus and species.
56. (b) Kidneys are not associated with the production of white blood cells. Kidneys regulate blood volume and composition, release erythropoietin and excrete waste in the urine. Bone marrow is involved in hematopoiesis. It is the site of Blymphocytes synthesis and maturation. Liver produces monocytes (a type of white blood cells). In spleen, B and Tlymphocytes are present. 50% of spleen cells are Blymphocytes and 3040% are Tlymphocytes.
57. (b) Hydathodes are involved in guttation. Hydathodes are specialised pore located along the leaf margins and tip which secrets water droplets. The exudation of water droplets from the tip or margin of the leaves is called guttation. Hydathodes mediated guttation occurs under high humidity and in the absence of transpiration. Cuticle is an extracellular layer which covers the epidermis of plants which provides protection against dessication and external environmental stress. Lenticels and stomata both regulates gaseous exchange between internal plant tissues and atmosphere and also regulates water movement through transpiration.
58. (d) Cataract affect the lens in eye. It occurs due to the clouding of lens and prevent light and image from reaching to retina. Cataract makes a person vision blurry and less colourful.
59. (a) Liverwort These are nonvascular plants and one of the three ancient lines of bryophytes (liverworts, hornworts and mosses). Volvox It is a spherical multicellular green algae and used as a genetic model of morphogenesis. Chlamydomonas It is a genus of unicellular green algae found in soil, freshwater and oceans. Fern These are vascular plants that possess true roots, leaves and stem and are reproduced by spores. Ferns and lycophytes are pteridophytes.
13 60. (d) The second meiotic division occurs after fertilisation. Oogenesis is the formation of female gametes (egg). Oogenesis begins in female before birth. During early fetal development, germ cell differentiate into oogonia. After several mitotic divisions, oogonia begins meiosis and known as primary oocytes. It remains arrested after diplotene of prophaseI of meiosisI until the female becomes sexually mature. After puberty, primary oocyte completes meiosisI and produces secondary oocytes and it arrests at metaphaseII and it completes meiosis II only after fertilisation. 61. (c) It is given that the quadratic equation x2 − 5cx − 6d = 0 has roots a and b, then …(i) a + b = 5c and ab = − 6d …(ii) and, the quadratic equation x2 − 5ax − 6b = 0 has roots c and d, then …(iii) c + d = 5a and …(iv) cd = − 6b Now, from Eqs. (i) and (iii), we have (a + b) − (c + d ) = 5c − 5a ⇒ (a − c) + (b − d ) = − 5(a − c) …(v) ⇒ (b − d ) = 6(c − a ) ∴a and c are the roots of equations. x2 − 5cx − 6d = 0 and x2 − 5ax − 6b = 0, respectively. ∴ a 2 − 5ac − 6d = 0 and c2 − 5ac − 6b = 0 ⇒ (a 2 − c2 ) − 6(d − b) = 0 6(d − b) …(vi) = 36 ⇒ a+ c= a−c From Eqs. (i) and (iii), we have (a + b) + (c + d ) = 5(a + c) ⇒ b + d = 4(a + c) = 4(36) [from Eq. (vi)] ⇒
b + d = 144
62. (a) The quadratic equation 4x2 + bx + c = 0 has equal roots if b2 − 16c = 0 ⇒ b2 = 24 c Now,c should be chosen from the set S = {1, 2, 3, … , 100}, such that it is a perfect square number, so c = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 ∴number of ordered pair (b, c) will be 10. 10 So, required probability = 100 × 100 1 = = 0.001 1000
KVPY Question Paper 2019 Stream : SA
14 63. (a) It is given that for n ∈ N fn = (n + 1)1/3 − n1/3 (n + 1) − n = / / / / n 23 (n + 1)23 + (n + 1)23 + n 23 1 = / / / / (n + 1)23 + (n + 1)23 n 23 + n 23 Q∀ n ∈ N
⇒
/ / / / / 3n 23 < (n + 1)23 + (n + 1)23 n 23 + n 23 / < 3(n + 1)23 1
3(n + 1)
23 /
1
1. If x, y are arbitrary real numbers such that x2 + y2 ≤ 1, then the minimum value of (x − x0 )2 + ( y − y0 )2 is
V0
0°C
V0
10°C
0°C
10°C
17. A very large block of ice of the size of a volleyball court and of uniform thickness of 8 m is floating on water. A person standing near its edge wishes to fetch a bucketful of water using a rope. The smallest length of rope required for this is about (a) 3.6 m
(b) 1.8 m
(c) 0.9 m
(d) 0.4 m
18. A box filled with water has a small hole on its side near the bottom. It is dropped from the top of a tower. As it falls, a camera attached on the side of the box records the shape of the water stream coming out of the hole. The resulting video will show (a) the water coming down forming a parabolic stream (b) the water going up forming a parabolic stream (c) the water coming out in a straight line (d) no water coming out
19. An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries 4 kg of water and the rate of evaporation is 20 g per hour, temperature of water in it decreases by ∆T in two hours. The value of ∆T is close to (ratio of latent of evaporation to specific heat of water is 540°C) (a) 2.7°C
(b) 4.2°C
(c) 5.4°C
(d) 10.8°C
20. Two plane mirrors are kept on a horizontal table
making an angle θ with each other as shown schematically in the figure. The angle θ is such that any ray of light reflected after striking both the mirrors returns parallel to its incident path. For this to happen, the value of θ should be
15. All the vertices of a rectangle are of the form (a , b) with a , b integers satisfying the equation (a − 8)2 − (b − 7)2 = 5 . Then, the perimeter of the rectangle is (a) 20 (c) 24
(b) 22 (d) 26
θ
(a) 30°
(b) 45°
(c) 60°
(d) 90°
3
KVPY Question Paper 2018 Stream : SA 21. A certain liquid has a melting point of −50°C and a boiling point of 150°C. A thermometer is designed with this liquid and its melting and boiling points are designated at 0°L and 100°L. The melting and boiling points of water on this scale are (a) 25°L and 75°L, respectively (b) 0°L and 100°L, respectively (c) 20°L and 70°L, respectively (d) 30°L and 80°L, respectively
acquired by an αparticle when it is accelerated by a potential of 1 V. For this problem, you may take a proton to be 2000 times heavier than an electron. Then, (b) 1 αV = 2 eV (d) 1 αV = 1eV
23. In a particle accelerator, a current of 500 µA is carried by a proton beam in which each proton has a speed of 3 × 107 m/s. The crosssectional area of the beam is 1.50 mm2. The charge density in this beam (in C/m3 ) is close to (a) 10−8
(b) 10−7
(c) 10−6
(a) We can see a rainbow in the western sky in the late afternoon (b) The double rainbow has red on the inside and violet on the outside (c) A rainbow has an arc shape, since the earth is round (d) A rainbow on the moon is violet on the inside and red on the outside
28. Remote sensing satellites move in an orbit that is at
22. One can define an alphavolt (αV ) to be the energy
(a) 1 α V = 1eV/4000 (c) 1 αV = 8000 eV
27. Select the correct statement about rainbow.
an average height of about 500 km from the surface of the earth. The camera onboard one such satellite has a screen of area A on which the images captured by it are formed. If the focal length of the camera lens is 50 cm, then the terrestrial area that can be observed from the satellite is close to (a) 2 × 10 3 A
(b) 106 A
29. Letters A, B, C and D are written on a cardboard as shown in the figure below. A D
(d) 10−5
24. Which of the following is not true about the total lunar eclipse? (a) A lunar eclipse can occur on a new moon and full moon day (b) The lunar eclipse would occur roughly every month, if the orbits of earth and moon were perfectly coplanar (c) The moon appears red during the eclipse because the blue light is absorbed in earth’s atmosphere and red is transmitted (d) A lunar eclipse can occur only on a full moon day
The cardboard is kept at a suitable distance behind a transparent empty glass of cylindrical shape. If the glass is now filled with water, one sees an inverted image of the pattern on the cardboard when looking through the glass. Ignoring magnification effects, the image would appear as (a) B
(a) the current density decreases in value (b) the magnitude of the electric field increases (c) the current density remains constant (d) the average speed of the moving charges remains constant
A D
D
B C
C
(c)
(d)
C B
D
A B
D C
30. If a ball is thrown at a velocity of 45 m/s in vertical upward direction, then what would be the velocity profile as function of height? (Assume, g = 10 m/s 2) (a)
(b) 45
45
v(m/s)
v(m/s)
0
26. A steady current I is set up in a wire whose crosssectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region,
(b)
A
A
(a) the minimum observed intensity of the parent star is 0.9 I 0 (b) the minimum observed intensity of the parent star is 0.99 I 0 (c) the minimum observed intensity of the parent star is 0.999 I 0 (d) the minimum observed intensity of the parent star is 0.9999 I 0
B C
25. Many exoplanets have been discovered by the transit method, where in one monitors, a dip in the intensity of the parent star as the exoplanet moves in front of it. The exoplanet has a radius R and the parent star has radius 100 R. If I 0 is the intensity observed on earth due to the parent star, then as the exoplanet transits
(d) 4 × 1012 A
(c) 1012 A
0
0 Height 101 (m)
(c)
0 Height 101 (m)
(d) 45
45
v(m/s)
v(m/s)
0
0 Height 101 (m)
0
0 Height 101 (m)
KVPY Question Paper 2018 Stream : SA
4
37. The correct statement about the following compounds
CHEMISTRY 31. The number of water molecules in 250 mL of water is closest to [Given, density of water is 1.0 g mL−1; Avogadro’s number = 6023 . × 1023 ] (a) 83.6 × 1023 (c) 1.5 × 1023
(b) 13.9 × 1023 (d) 33.6 × 1023
32. Among the following, the correct statement is (a) pH decreases when solid ammonium chloride is added to a dilute aqueous solution of NH3 (b) pH decreases when solid sodium acetate is added to a dilute aqueous solution of acetic acid (c) pH decreases when solid NaCl is added to a dilute aqueous solution of NaOH (d) pH decreases when solid sodium oxalate is added to a dilute aqueous solution of oxalic acid
Br Y
Br X
is (a) Both are chiral (b) Both are achiral (c) X is chiral and Y is achiral (d) X is achiral and Y is chiral
38. The most acidic proton and the strongest nucleophilic nitrogen in the following compound O
a
33. The solubility of BaSO4 in pure water (in g L −1) is
b
c
N H
N H
CH3
N H
closest to [Given; Ksp for BaSO4 is 10 . × 10−10 at 25°C. Molecular −1
weight of BaSO4 is 233 g mol ] (a) 10 . × 10−5 (b) 10 . × 10−3 (c) 2. 3 × 10−5 (d) 2. 3 × 10−3
respectively, are (a) N a − H; Nb (c) N a − H; N c
(b) Nb − H; N c (d) N c − H; N a
39. The chlorine atom of the following compound Cl Cl d
34. Among the following, the incorrect statement is (a) no two electrons in an atom can have the same set of four quantum numbers (b) the maximum number of electrons in the shell with principal quantum number, n is equal to n 2 + 2 (c) electrons in an orbital must have opposite spin (d) in the ground state, atomic orbitals are filled in the order of their increasing energies
35. A container of volume 2.24 L can with stand a
Cl
O
Cl
b
a
that reacts most readily with AgNO3 to give a precipitate is (a) Cl a
(b) Clb
(c) Cl c
(d) Cl d
40. Among the following sets, the most stable ionic species are
maximum pressure of 2 atm at 298 K before exploding. The maximum amount of nitrogen (in g) that can be safely put in this container at this temperature is closest to
(a)
(a) 2.8 (c) 1.4
(c)
(b) 5.6 (d) 4.2
c
–
+
and
and
and
(b)
+
–
–
–
+
+
(d)
and
36. The compound shown below O
NO2
41. The correct order of energy of 2sorbitals in H, Li, Na and K, is (a) K < Na < Li < H (c) Na < K < H < Li
(b) Na < Li < K < H (d) H < Na < Li < K
42. The hybridisation of xenon atom in XeF4 is can be readily prepared by FriedelCraft’s reaction between (a) benzene and 2nitrobenzoyl chloride (b) benzyl chloride and nitrobenzene (c) nitrobenzene and benzoyl chloride (d) benzene and 2nitrobenzyl chloride
(a) sp3 (c) sp3 d 2
(b) dsp 2 (d) d 2sp3
43. The formal oxidation numbers of Cr and Cl in the ions Cr2O72− and ClO3− , respectively are (a) + 6 and +7 (c) +6 and +5
(b) +7 and +5 (d) +8 and +7
5
KVPY Question Paper 2018 Stream : SA 44. A filter paper soaked in salt X turns brown when exposed to HNO3 vapor. The salt X is (a) KCl
(b) KBr
(c) KI
53. Which one of the following proteins does not play a role in skeletal muscle contraction?
(d) K2SO4
45. The role of haemoglobin is to (a) store oxygen in muscles (b) transport oxygen to different parts of the body (c) convert CO to CO2 (d) convert CO2 into carbonic acid
(a) Actin (c) Troponin
(b) Myosin (d) Microtubule
54. Which one of the following reactions is catalysed by highenergy ultraviolet radiation in the stratosphere? (a) O2 + O → O3 (c) O3 + O3 → 3O2
(b) O2 → O + O (d) O + O → O2
55. Which one of the following statements is true about trypsinogen?
BIOLOGY 46. Which one of the following molecules is a secondary metabolite? (a) Ethanol (c) Penicillin
(b) Lactate (d) Citric acid (b) phospholipid (d) protein
48. The water potential (ψ p ) of pure water at standard temperature and atmospheric pressure is (a) 0
(b) 0.5
(c) 1.0
(d) 2.0
49. Action potential in neurons is generated by a rapid influx of (a) chloride ions (c) calcium ions
(b) potassium ions (d) sodium ions
50. Erythropoietin is produced by (a) heart (c) bone marrow
(b) kidney (d) adrenal gland
(c) It is activated by pepsin (d) It does not need activation
through the skin? (a) Blue whale (c) Platypus
(b) Salamander (d) Peacock
57. Which one of the following human cells lacks a nucleus? (a) Neutrophil (c) Mature erythrocyte
(b) Neuron (d) Keratinocyte
58. The first enzyme that the food encounters in human digestive system is (a) pepsin
(b) trypsin
(c) chymotrypsin
(d) amylase
59. Glycoproteins are formed in which one of the following organelles?
51. Tendrils are modifications of (a) stem or leaf (c) leaf only
(b) It is activated by renin
56. Which one of the following organisms respires
47. Lecithin is a (a) carbohydrate (c) nucleoside
(a) It is activated by enterokinase
(b) stem only (d) aerial roots only
52. Which one of the following combinations of
(a) Peroxisome (c) Golgi apparatus
(b) Lysosome (d) Mitochondria
60. An example of nastic movement (external
biomolecules is present in the ribosomes?
stimulusdependent movement) in plants is
(a) RNA, DNA and protein (b) RNA, lipids and DNA (c) RNA and protein (d) RNA and DNA
(a) folding up of the leaves of Mimosa pudica (b) climbing of tendrils (c) growth of roots from seeds (d) growth of pollen tube towards the ovule
PARTII (2 Marks Questions) MATHEMATICS
(a) 9
(b) 12
(c) 15
(d) 18
61. What is the sum of all natural numbers n such that
63. The number of solid cones with integer radius and
the product of the digits of n (in base 10) is equal to n 2 − 10n − 36?
integer height each having its volume numerically equal to its total surface area is
(a) 12
(a) 0
(b) 13
(c) 124
(d) 2612
62. Let m (respectively, n) be the number of 5digit integers obtained by using the digits 1, 2, 3, 4, 5 with repetitions (respectively, without repetitions) such that the sum of any two adjacent digits is odd. Then m is equal to n
(b) 1
(c) 2
(d) infinite
64. Let ABCD be a square. An arc of a circle with A as centre and AB as radius is drawn inside the square joining the points B and D. Points P on AB, S on AD, Q and R on arc BD are taken such that PQRS is a square.
KVPY Question Paper 2018 Stream : SA
6 Further suppose that PQ and RS are parallel to AC. area PQRS is Then, area ABCD (a)
1 8
(b)
1 5
(c)
1 4
(d)
(a) T
(b) T
2 5
65. Suppose ABCD is a trapezium whose sides and height are integers and AB is parallel to CD. If the area of ABCD is 12 and the sides are distinct, then  AB − CD (a) is 2 (b) is 4 (c) is 8 (d) cannot be determined from the data
E (c) T
E (d) T
E
E
CHEMISTRY 71. Among the following, the species with identical bond order are
PHYSICS 66. A coffee maker makes coffee by passing steam through a mixture of coffee powder, milk and water. If the steam is mixed at the rate of 50 g per minute in a mug containing 500 g of mixture, then it takes about t0 seconds to make coffee at 70° C when the initial temperature of the mixture is 25°C. The value of t0 is close to (ratio of latent heat of evaporation to specific heat of water is 540°C and specific heat of the mixture can be taken to be the same as that of water) (a) 30
(b) 45
(c) 60
(d) 90
67. A person in front of a mountain is beating a drum at the rate of 40 per minute and hears no distinct echo. If the person moves 90 m closer to the mountain, he has to beat the drum at 60 per minute to not hear any distinct echo. The speed of sound is (a) 320 ms−1 (b) 340 ms−1 (c) 360 ms−1 (d) 380 ms−1
68. A glass beaker is filled with water up to 5 cm. It is kept on top of a 2 cm thick glass slab. When a coin at the bottom of the glass slab is viewed at the normal incidence from above the beaker, its apparent depth from the water surface is d cm. Value of d is close to (the refractive indices of water and glass are 1.33 and 1.5, respectively) (a) 2.5 cm
(b) 5.1 cm
(c) 3.7 cm
(d) 6.0 cm
69. A proton of mass m and charge e is projected from a
very large distance towards an αparticle with velocity v. Initially αparticle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be (a) e2/4 π ε0 mv2 (c) 2e2/4 π ε0 mv2
(b) 5e2/4 π ε0 mv2 (d) 4e2/4 π ε0 mv2
70. A potential is given by V (x) = k(x + a )2 / 2 for x < 0 and V (x) = k(x − a )2 / 2 for x > 0. The schematic variation of oscillation period T for a particle performing periodic motion in this potential as a function of its energy E is
(a) CO and O2− 2 (c) O2− 2 and B2
(b) O−2 and CO (d) CO and N+2
72. The quantity of heat (in J) required to raise the temperature of 1.0 kg of ethanol from 293.45 K to the boiling point and then change the liquid to vapor at that temperature is closest to [Given, boiling point of ethanol 351.45 K. Specific heat capacity of liquid ethanol 2.44 J g −1K −1. Latent heat of vaporisation of ethanol 855 J g −1] (a) 142 . × 102 (c) 142 . × 105
(b) 9.97 × 102 (d) 9.97 × 105
73. A solution of 20.2 g of 1,2dibromopropane in MeOH upon heating with excess Zn produces 3.58 g of an unsaturated compound X. The yield (%) of X is closest to [Atomic weight of Br is 80.] (a) 18
(b) 85
(c) 89
(d) 30
74. The lower stability of ethyl anion compared to methyl anion and the higher stability of ethyl radical compared to methyl radical, respectively, are due to (a) + Ieffect of the methyl group in ethyl anion σ → porbital conjugation in ethyl radical (b) − Ieffect of the methyl group in ethyl anion and σ → σ * conjugation in ethyl radical (c) + I effect of the methyl group in both cases (d) + I effect of the methyl group in ethyl anion and σ → σ * conjugation in ethyl radical
75. The FBrF bond angles in BrF5 and the ClPCl bond angles in PCl5 , respectively, are (a) identical in BrF5 but nonidentical in PCl5 (b) identical in BrF5 and identical in PCl5 (c) nonidentical in BrF5 but identical in PCl5 (d) nonidentical in BrF5 and nonidentical in PCl5
BIOLOGY 76. If the genotypes determining the blood groups of a couple are IA IO and IA IB , then the probability of their first child having type O blood is (a) 0
(b) 0.25
(c) 0.50
(d) 0.75
7
KVPY Question Paper 2018 Stream : SA 77. A cross was carried out between two individuals
Column I
heterozygous for two pairs of genes was carried out. Assuming segregation and independent assortment, the number of different genotypes and phenotypes obtained respectively would be (a) 4 and 9 (c) 9 and 4
(b) 6 and 3 (d) 11 and 4
Column II
P. Hypermetropia
(i) Nearsightedness a. Convex lens
Q. Myopia
(ii) Farsightedness
(a) P–ii–b (c) P–i–a
78. If the H+ concentration of an aqueous solution is
Column III
b. Concave lens
(b) Q–i–b (d) Q–i–a
80. Which one of the following properties causes the
0.001 M, then the pOH of the solution would be
plant tendrils to coil around a bamboo stick?
(a) 0.001
(a) Tendril has spines (b) The base of the tendril grows faster than the tip (c) Part of the tendril in contact with the bamboo stick grows at a slower rate than the part away from it. (d) The tip of the tendril grows faster than the base
(b) 0.999
(c) 3
(d) 11
79. Consider the following vision defects listed in Columns I and II and the corrective measures in Column III. Choose the correct combination.
Answers PARTI 1
(d)
2
(b)
3
(c)
4
(b)
5
(a)
6
(b)
7
(c)
8
(c)
9
(c)
10
(a)
11
(c)
12
(d)
13
(b)
14
(a)
15
(a)
16
(a)
17
(c)
18
(d)
19
(c)
20
(d)
21
(a)
22
(b)
23
(d)
24
(a)
25
(d)
26
(b)
27
(*)
28
(c)
29
(d)
30
(a)
31
(a)
32
(a)
33
(d)
34
(b)
35
(d)
36
(a)
37
(c)
38
(b)
39
(a)
40
(d)
41
(a)
42
(c)
43
(c)
44
(c)
45
(b)
46
(c)
47
(b)
48
(a)
49
(d)
50
(b)
51
(a)
52
(c)
53
(d)
54
(b)
55
(a)
56
(b)
57
(c)
58
(d)
59
(c)
60
(a)
PARTII 61
(b)
62
(c)
63
(b)
64
(d)
65
(b)
66
(b)
67
(c)
68
(b)
69
(b)
70
(b)
71
(c)
72
(d)
73
(b)
74
(a)
75
(d)
76
(a)
77
(c)
78
(d)
79
(b)
80
(c)
* No options are correct.
Solutions 1. (d) We have,
2. (b) Given,
a 4 + b4 < 1 and a 2 + b2 > 1 The graph of x2 + y2 = 1 and x4 + y4 = 1 are
x4 − x2 + 2x − 1 = 0 ⇒ x4 − (x − 1)2 = 0 2 ⇒ (x − x + 1) (x2 + x − 1) = 0 ⇒ x2 − x + 1 = 0 or x2 + x − 1 = 0 ⇒ x2 − x + 1 = 0 has no real roots. ⇒ x2 + x − 1 = 0 has two real roots
Y x4+y4=1
X′
X
3. (c) Given, Sm = n and Sn = m m [2a + (m − 1)d ] = n ...(i) 2 n Sn = (2a + (n − 1) d ) = m ...(ii) 2 On subtracting Eq. (ii) from Eq. (i), we get d (m − n ) a + (m − n ) (m + n − 1) 2 Sm =
x2+y2=1 Y′
Clearly from graph. There are many positive real number (a , b) satisfying a 4 + b4 < 1 and a 2 + b2 > 1.
= − (m − n ) ⇒ ∴
2a + (m + n − 1) d = − 2 [m ≠ n ] m+ n Sm + n = (2a + (m + n − 1) d ) 2 m+ n (− 2) = − (m + n ) = 2
4. (b) (I) Any pair of consistent linear equation in two variables must have a unique solution. This statement is false. Consistent equation may have unique or infinite solution. (II) There do not exists two consecutive integers the sum of whose square is 365. This statement is also false 132 + 142 = 365.
KVPY Question Paper 2018 Stream : SA
8 5. (a) Let P (x) = anx + an − 1 x n
n −1
+ an − 2x
n−2
+ ... + a1 x + a0
a0 , a1 , a2 K ∈ I Given, P (2) = 2 and P (4) = 5 2 = an 2n + an − 1 2n − 1 + an − 2 2n − 2 + ... + a1 2 + a0 … (i) 5 = an 4n + an − 1 4n − 1 + an− 2 4n − 2 + .... + 4a1 + a0 … (ii) On subtracting Eq. (i) from Eq. (ii), we get 3 = an (4n − 2n ) + an − 1 (4n − 1 − 2n − 1 ) + ... + 2a1 Clearly, LHS is odd number and RHS is even number. ∴No polynomials exists.
6. (b) Four digits number which is divisible by 7 are 1001, 1008, 1015, ...., 9996. Hence, total number of such numbers = 1286 N 2 Median =
th
2 3 πr 3 When height of cylinder is doubled, then 2 volume of solid = 2 πr 2h + πr3 3 2 2 πr 2h + πr3 V2 3 3 = = ∴ 2 V1 2 πr 2h + πr3 3 2 2h + r 3 h r 3 = ⇒ ⇒ = 2 2 3 2 h+ r 3 When the radius is doubled, then volume 16 πr3 of solid = 4 πr 2h + 3 16 4h + r V′ 3 = 4h + 8h = 6 Q r = h ∴ 2= 3 2 2 h+ h V1 h+ r 3 Hence, volume is increased by 500%. ∴Volume of solid = πr 2h +
th
observation
Given, QR 2 + PR 2 = 5PQ 2 Median PM and QN intersect of G. P
2 N
[Q N is even] 1286 2 Median =
2 3 πr 3
8. (c) In ∆PQR
observation
N + + 1 2
and volume of hemisphere =
th
G
observation
1286 + + 1 2
th
2 643th + 644th = 2 (1001 + (642)7) + (1001) + (643)7) = 2 2 (1001) + 7 (642 + 643) = 2 2 (1001) + 7 (1285) = 2 = 1001 + 4497.5 = 5498.5
7. (c) Let the height and radius of cylinder are h are r, respectively. ∴Volume of cylinder = πr 2h r
r
h
h r
Q
observation
M
QG = ⇒
2 1 QN , GM = PM 3 3 2
R
2
2 1 QG 2 + GM 2 = QN + PM 3 3 4 1 = QN 2 + PM 2 9 9 4 2PQ 2 + 2QR 2 − PR 2 = 9 4 2 1 2PQ + 2PR 2 − QR 2 + 9 4 8PQ 2 + 8QR 2 − 4PR 2 1 + 2PQ 2 + 2PR 2 − QR 2 = 9 4 1 10PQ 2 + 7QR 2 − 2PR 2 = 9 4 2 (5PQ 2 − PR 2 ) + 7QR 2 4 2 2 1 2QR + 7QR 1 2 2 = = QR = QM 9 4 4 =
∴ ∴
OG 2 + GM 2 = QM 2 ∠QGM = 90°
9. (c) Let ∆ABC BC = a AC = b AB = c A E
F G B
C
D
and median of ∆ABC AD = l BE = m CF = n AD is median, AB + BC AD < ∴ 2 b+ c ∴ l< 2 a+ b a+ c and n < Similarly, m< 2 2 ∴ l+ m+ n< a+ b+ c l+ m+ n …(i) BC 2 (m + n ) > a ∴ 3 2 2 Similarly, (n + l) > b and (m + l) > c 3 3 4 (l + m + n ) > a + b + c Q 3 l+ m+ n 3 …(ii) ⇒ > a+ b+ c 4 From Eqs. (i) and (ii), we get l + m+ n 3 ∈ , 1 a + b+ c 4
10. (a) Let P (x0 , y0 ) Given x2 + y2 ≤ 1 Let any arbitrary point 8(x, y). Y (0,1) Q
P(x0,y0) (x,y)
1 X′
O
1 9
Y′
(1,0)
X
9
KVPY Question Paper 2018 Stream : SA PQ 2 = (x − x0 )2 + ( y − y0 )2 PQ 2 = (OP − OQ )2 PQ 2 = (OP − OQ )2 PQ 2 = ( x02 + x02 − 1)2
14. (a) Given, PQR is an acute angle
16. (a) As temperature of water is
triangle.
increased from 0°C to 10°C, density of water initially increases upto a maximum at 4°C and then it reduces. So, buoyant force on block of wood also increases till temperature reaches 4°C and then decreases from 4°C to 10°C. Hence, volume of block above water also increases upto 4°C and then decreases from 4°C to 10°C. ∴Variation of V 0 versus t as shown below.
PQ < QR [Q OQ = 1]
Q
∴Mininimum value of PQ is 2
p r
( x02 + y02 − 1)2
11. (c) Given, in ∆PQR Altitude ⇒
PQ = 3 RS = 3 PS = QR
P
Q1
Q2 q
R
Q3
∠QRP < ∠QPR 1 PQ3 = PR 2 PQ2 : Q2 R = r : p r PQ2 = PR r + p
R
V0
0°C 4°C
r< p 1 PQ2 < PR 2 Comparison between altitude and angle bisector ∠QPQ2 + ∠PQ2Q + ∠PQQ2 = ∠RQQ2 + ∠QQ2R + ∠QRQ2 ∴∠PQQ2 = ∠RQQ2 [since, QQ2 is angle bisector of ∠Q] ∠QPQ2 + ∠PQ2Q = ∠QQ2R + ∠QRQ2 ∴PQ < QR the ∠QPQ2 < ∠QRQ2 Hence, ∠QQ2P < ∠QQ2R But ∠QQ2P + ∠QQ2R = 180° Hence, ∠QQ2P < 90° and ∠QQ2R > 90° Q Foot from Q to side PR lie inside ∆PQQ2 ⇒ PQ1 < PQ2 < PQ3 But
P
S
Q
In ∆SQR, QR 2 = SR 2 + SQ 2 PS 2 = ( 3 )2 + (QP − PS)2 [Q SQ = PQ − PS] PS 2 = 3 + (3 − PS)2 PS 2 = 3 + 9 − 6PS + PS 2 ⇒ PS = 2 In ∆PRS, PR 2 = PS 2 + RS 2 = (2)2 + ( 3 )2 = 4 + 3 PR =
7
12. (d) Total number of students = 50 Average marks of student = 47.5 ∴Total marks of students = 50 × 47.5 = 2375 Now, the student get integer marks Hence, the maximum number of students we will divide total mark by 48. 2375 = 49 ∴ 48
13. (b) Given number, n = 152 × 518 n = 32 × 52 × 518 n = 9 × 520 Taking log base 10 both side log10 n = log10 9 + log10 520 = 2 log10 3 + 20 log10 5 = 2 × 0.4771 + 20 × (1 − 0.3010) = 14 characters value Hence, the number have 15 digits S = Sum of digits of the number Now, n has last digit is 5. ∴Minimum value of S = 1 + 5 = 6 Maximum value of S = 9 × 14 + 5 = 126 + 5 = 131 ∴ 6 ≤ S < 140
15. (a) Given, (a − 8)2 − (b − 7)2 = 5 ⇒ (a − 8 + b − 7) (a − 8 − b + 7) = 5 ⇒ (a + b − 15) (a − b − 1) = 5 There are four case … (i) a + b − 15 = 5; a − b − 1 = 1 ... (ii) a + b − 15 = 1; a − b − 1 = 5 a + b − 15 = − 5; a − b − 1 = − 1 … (iii) ... (iv) a + b − 15 = − 1; a − b − 1 = − 5 On solving, we get (i) a = 11, b = 9 (ii) a = 11, b = 5 (iii) a = 5, b = 5 (iv) a = 5, b = 9 D(11, 5)
4
6
A(5, 5)
C(11, 5)
6
4
∴Perimeter = 2(4 + 6) = 20
B(5, 5)
10°C
t
17. (c) Fraction of thickness of ice block out of water is ρ 0.9 or x = 01 x = 1 − ice = 1 − . 1 ρwater
0.8 m 8m
So, minimum length of rope required ≈ thickness of ice × 01 . = 8 × 01 . = 0.8 m. Hence, nearest option is 0.9 m.
18. (d) When box with hole is in free fall, both water and box cover equal distance downwards in equal time. Hence, no water comes out of hole in free fall of box.
19. (c) Water evaporated in two hours = m = 2 h × 20 g/h = 40 g = 40 × 10−3 kg Heat absorbed by water during evaporation is Q = Mass evaporated × Latent heat …(i) Q = mL Assuming this heat is taken entirely from water in earthen pot, if ∆T is decrease of temperature of pot then, …(ii) Q = Ms∆T where, M = mass of water in pot and s = specific heat of water. Equating Eqs. (i) and (ii), we get mL = Ms∆T m L 40 × 10−3 or ∆T = × = × 540 = 5.4°C 4 M s
KVPY Question Paper 2018 Stream : SA
10 20. (d) As emergent ray is parallel to
24. (a) A lunar eclipse occurs only on a
27. (No option is matching)
incident ray, deviation angle δ is 180°. But δ = 360°− 2θ where, θ = angle between inclined mirrors.
full moon day.
In late afternoon rainbow is visible in east side when light of sun in west side is reflected and refracted by a layer of water droplets. Rainbow is circular because locous of reflected rays reaching eye of observer is a circle. Its roundness is not due to roundness of earth. There is no rainbow on moon due to lack of atmosphere. In case of a primary rainbow, violet colour is on inside and red colour is on outside of arc. In case of a secondary rainbow, red colour is on inside and violet colour is on outside of arc.
So, option (a) is incorrect.
25. (d) Intensity of radiation (mainly visible light) emitted from surface of a star is proportional to its area. I ∝ A or I = kA
So,
where, k = constant. Now, if I 0 = intensity of parent star.
θ
So, or
Then, I 0 = kπ (100 R )2 = k π R 2 × 10000
360° − 2θ = 180° 2θ = 180° ⇒ θ = 90°
21. (a) From principle of thermometry,
T − TLFP = a constant for every TUFP − TLFP thermometric scale. Now, for any temperature L on a thermometer designed with given liquid and equivalent temperature C on centigrade scale, we have C − TLFP L − TLFP Liquid = TUFP − TLFP Centigrade TUFP − TLFP based scale
22. (b) An alphavolt (αV ) is the energy acquired by an αparticle (charge 2e units) when accelerated by a potential difference of 1 V.
R
K DAR
Exoplanet radius R
PRIM
⇒
500 × 10−6
ARY
RAIN
W BO
R
I min = kπR 2 (10000 − 1) = kπR × 9999
So,
I min kπR 2 × 9999 = I0 kπR 2 × 10000
⇒
I min = I 0 × 0.9999
V
V
42° 51°
2
Observer East West
26. (b) When current flows through a conductor of tapered crosssection, current flow through every section remains constant.
So, none of the option is correct. Option (b) is correct, if only secondary rainbow is considered.
28. (c) Consider the given diagram, Screen
I
⇒ ⇒ ⇒
A1
A2
I
I1 = I 2 j1 A1 = j2A2 j1 A2 = h AB + CD = 24, 12, 8, 6
B
KVPY Question Paper 2018 Stream : SA
14 In ∆BEC, BEC is a right angled triangle. ∴h must be 3 and 4 When h = 3, BE = 4, CE = 5 AB + CD = 8 AE + BE + AE = 8 2AE = 8 − BE = 8 − 4 AE = 2 ∴ AB = 4 + 2 = 6, CD = 2 ∴ AB − CD = 6 − 2 = 4
66. (b) Let m gram of steam is condensed in the process of heating mixture from 25°C to 70°C. Then, Heat lost by steam = Heat gained by mixture ⇒ Heat of condensation of steam + Heat given by water formed = Heat gained by mixture ⇒ m ⋅ L + msw ∆T = M ⋅ sm∆T ⇒ mL + msw (100 − 70) = 500 × sw × (70 − 25) 500 × sw × 45 ⇒m = L + 30 sw 500 × 45 ⇒m = L + 30 sw 500 × 45 ⇒ m= ≈ 40 g (540 + 30) Now, in 1 min, 50 g of steam is condensed. ∴ 40 g of steam will be condensed in time t0 , 40 × 60 s = 48 s t0 = 50 Nearest answer is 45 s.
67. (c) As drummer does not hear any echo this means time between two successive wavefronts is equal to time in which a wavefront reaches back to drummer. x
Distance covered by sound = 2x If v = speed of sound, then 2x = time interval between two successive v wavefronts.
So, we have 2x 60 In case I, …(i) = v 40 2(x − 90) 60 In case II, …(ii) = v 60 Substituting for x from Eq. (i) in Eq. (ii), we get 2x − 180 = v 3 v − 180 = v ⇒ 2 ⇒ v = 360 ms−1 68. (b) Apparent depth d in case of more than one medium is d d …(i) d = 1 + 2 + ... µ1 µ 2 where, d1 and d2 are the thickness of slabs of medium with refractive index µ 1 and µ 2, respectively. Here, d1 = 5 cm, µ 1 = 133 . d2 = 2 cm, µ 2 = 15 . Substituting these values in Eq. (i), we get 5 2 Apparent depth, d = + 133 . 15 . = 5.088 cm = 51 . cm
69. (b) As αparticle is free to move, initial kinetic energy of system will be 1 ki = µv2 2 where, µ = reduced mass of system m ⋅ 4m . = m + 4m Now, by energy conservation, we have Initial kinetic energy = Potential energy at minimum separation r 1 m . 4m 2 1 2e2 . v = 2 m + 4m 4 πε0 r ⇒
r=
5e2 4 πε0 mv2
70. (b) Given, potential function for the oscillating particle is k (x + a ) 2 , x< 0 2 V (x ) = 2 k (x − a ) , x > 0 2 So, potential energy of the particle (mass m) is km (x + a )2 , x< 0 2 U (x ) = 2 km(x − a ) , x < 0 2 dU km(x + a ), x < 0 = dx km (x − a ), x > 0
dU = 0, when x = ± a. dx d 2U Now, = km> 0 dx2 So, particle is in unstable equilibrium at x = ± a. Hence, particle is unbounded for − a > x and x > a. In region, − a ≤ x ≤ a, time period of particle reduces from a maximum. So, correct graph is (b). If
71. (c) The bond order can be calculated as 1 (Nb − N a ) 2 where, Nb = electrons in bonding orbitals N a = electrons in antibonding orbitals. (a) CO and O2− 2 The electronic configuration of CO (14) is B.O =
σ1s2 σ * 1s2 σ 2s2 σ * 2s2 σ * 2 pz2 π 2 px2 π 2 p 2 y 1 6 B.O = (10 − 4) = = 3 ∴ 2 2 The electronic configuration of O22− (18) is σ1s2 σ * 1s2 σ 2s2 σ * 2s2σ 2 pz2 π 2 px2 π 2 py2 π * 2 px2 π * 2 py2 1 B.O = (10 − 8) = 1 2 − (b) O2 and CO The electronic configuration of O−2 (17) is σ1s2 σ * 1s2 σ 2s2 σ * 2s2 σ 2 pz2 π 2 px2 π 2 py2 π * 2 px2 π * 2 p1y 1 3 B.O = (10 − 7) = = 1.5 2 2 B.O of CO is 3 [as calculated in option (a)] (c) B.O of O2− is 1 2 [as calculated in option (a)] The electronic configuration of B2 (10) is σ1s2 σ * 1s2 σ 2s2 σ * 2s2 π 2 p1x π 2 p1y 1 2 B.O = [6 − 4] = = 1 2 2 (d) B.O of CO is 3 [as calculated in option (a)] Electronic configuration of N+2 (13) is σ1s2 σ * 1s2 σ 2s2 σ * 2s2 π 2 px2 π 2 py2 σ 2 p1z 1 5 B.O = [9 − 4] = = 2 .5 2 2 Thus, option (c) is correct.
72. (d) Given, mass of ethanol = 1kg = 1000 g Latent heat of vaporisation of ethanol = 855 Jg −1
15
KVPY Question Paper 2018 Stream : SA Specific heat capacity of ethanol = 2. 44 J / gk −1 Heat, q = mc∆T + heat of vaporisation = 1000 × 2 .44 (351.45 − 293.45) + 855 × 1000 J 5 = 9.97 × 10 J
The geometry of PCl5 is triangular pyramidal.
73. (b)
The axial bonds suffer more repulsions than equitorial bonds, so they are larger in bond length.
Br Zn
Br
MeOH 1, 2dibromopropane
Prop  1 ene (X)
Yr
Cl Cl Cl
yR
P Cl Cl
76. (a) The genotype of child having A O
Moles of 1, 2dibromo propane 20.2 . mole = = 01 202 3.58 Moles of prop1ene = 42 = 0.085 mole 0.085 % yield = × 100 = 85%. 01 . The lower stability of ethyl anion 74. (a) − (CH3 CH2 ) compared to methyl anion (CH3− ) is because of + Ieffect of methyl group of ethyl anion. The higher stability of ethyl radical compared to methyl radical is due to σ − porbital conjugation which is known as hyper conjugation in ethyl radical. H H
C
C
H H
H C
H
CH2
H
H
Hyper conjugation in ethyl radical
75. (d) The geometry of BrF5 is square pyramidal. F
blood groupO with parents having I I and IA IB blood groups can be represented as Parents IA IO
IA IB
×
YyRR
YyRr
yyRR
yellow round
yellow round
green green round round
YyRr
Yyrr
yyRr
yellow round
green wrinkled yellow wrinkled round green
yyRr
yyrr
The genotypic ratio is 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2:1 The phenotypic ratio is 9 : 3 : 3 : 1 ∴ The number of different genotypes and phenotypes obtained would be 9 and 4, respectively.
77. (c) In the given question, both
79. (b) Hypermetropia is far
Phenotype
IA IB
IA IO
IB IO
Blood Blood Blood Blood group –A group –AB group –A group –B
Offsprings
parents are heterozygous for two pairs of genes. This means the cross is a dihybrid cross. Lets assume a dihybrid cross Pure breeding traits Heterozygous trait
Yellow round , Wrinkled green seeds seeds (YYRR) (yyrr) Yellow round seeds (YyRr)
YR
Br
Yr
yR
yr
F
Here, the lone pair occupies the axial position and hence axial bonds will suffer more repulsion than axial bonds. Thus, the axial Br—F bond length will be different than equitorial Br—F.
yellow yellow yellow wrinkled round wrinkled
From the above cross, it is shown that none of the offsprings will be of blood groupO. ∴ The probability of their first child having typeO blood is zero.
Genotype IA IA
Gametes
F
Yyrr
YYrr
yellow round
78. (d) The H+ ion concentration of an aqueous solution is 0.001 M or 1 × 10−3 M Since we know that pH = − log[H+ ] Using this equation, by plugging in the values pH = − log 10−3 = − (− 3) log 10 = 3 pH = 3 We know that pOH = 14 − pH = 14 − 3 = 11 ∴ pOH = 11
F
F
yr
YyRr
YYRr
YR
Yr
yR
yr
YYRR
YYRr
YyRR
YyRr
yellow round
yellow round
yellow yellow round round
+
YR
sightedness. A vision condition in which nearby objects are blurry. It is corrected by using convex lens. Myopia is near sightedness. A condition in which close objects appear clearly but far ones do not. It is corrected using concave lens.
80. (c) The tendrils are sensitive to touch. When they come in contact with any support, the part of the tendril in contact with the object does not grow rapidly as the part of the tendril away from the object. This causes the tendril to circle around the object and thus cling to it. This process is known as positive thigmotropism. Thigmotropism is the directional response of a plant organ to touch or physical contact with a solid object. This differential response is generally caused by the induction of some pattern differential growth.
KVPY Question Paper 2017 Stream : SA
16
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2017 Stream : SA (Nov 19) MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS 1. Suppose BC is a given line segment in the plane and T is a scalene triangle. The number of points A in the plane such that the triangle with vertices A, B, C (in same order) is similar to triangle T is (a) 4
(b) 6
(c) 12
(d) 24
2. The number of positive integers n in the set
1 {2, 3, ... , 200} such that has a terminating decimal n expansion is (a) 16
(b) 18
(c) 40
(d) 100
3. If a , b, c are real numbers such that a + b + c = 0 and a 2 + b2 + c2 = 1, then (3a + 5b − 8c)2 + (− 8a + 3b + 5c)2 + (5a − 8b + 3c)2 is equal to (a) 49 (c) 147
(b) 98 (d) 294
4. Let ABC be a triangle and M be a point on side AC closer to vertex C than A. Let N be a point on side AB such that MN is parallel to BC and let P be a point
on side BC such that MP is parallel to AB. If the area 5 of the quadrilateral BNMP is equal to of the area 18 of ∆ ABC, then the ratio AM / MC equals (a) 5 18 (c) 5
(b) 6 15 2
(d)
5. Let n ≥ 4 be a positive integer and let l1 , l2 , ... , ln be the lengths of the sides of arbitrary n sided nondegenerate polygon P. Suppose ln − 1 ln l1 l2 + + ... + + =n l2 l3 ln l1 Consider the following statements: I. The lengths of the sides of P are equal. II. The angles of P are equal. III. P is a regular polygon if it is cyclic. Then, (a) I is true and I implies II (b) II is true (c) III is false (d) I and III are true
17
KVPY Question Paper 2017 Stream : SA 6. Consider the following statements: For any integer n,
(a) both I and II are true (b) both I and II are false (c) I is false and II is true (d) I is true and II is false
7. Let S be the set of all ordered pairs (x, y) of positive integers, with HCF (x, y) = 16 and LCM (x, y) = 48000. The number of elements in S is (a) 4
(b) 8
(c) 16
(d) 32
8. Consider the set A of natural numbers n whose units digit is nonzero, such that if this units digit is erased, then the resulting number divides n. If K is the number of elements in the set A, then (a) K is infinite (c) 25 ≤ K ≤ 10
(b) K is infinite but K > 100 (d) K < 25
9. There are exactly twelve Sundays in the period from January 1 to March 31 in a certain year. Then, the day corresponding to February 15 in that year is (a) Tuesday (b) Wednesday (c) Thursday (d) not possible to determine from the given data
10. Consider a threedigit number with the following
properties: I. If its digits in units place and tens place are interchanged, the number increases by 36; II. If its digits in units place and hundreds place are interchanged, the number decreases by 198. Now, suppose that the digits in tens place and hundreds place are interchanged. Then, the number (a) increases by 180 (c) increases by 360
(b) decreases by 270 (d) decreases by 540
11. Consider four triangles having sides (5, 12, 9), (5, 12, 11), (5, 12, 13) and (5, 12, 15). Among these, the triangle having maximum area has sides. (a) (5,12, 9) (c) (5, 12, 13)
(b) (5, 12, 11) (d) (5, 12, 15)
12. In a classroom, onefifth of the boys leave the class and the ratio of the remaining boys to girls is 2 : 3. If further 44 girls leave the class, then class the ratio of boys to girls is 5 : 2. How many more boys should leave the class so that the number of boys equals that of girls? (a) 16
(b) 24
(c) 30
14. The least value of a natural number n such that n n − 1 n − 1 n n! , is + < , where = r (n − r )! r ! 5 6 7
I. n 2 + 3 is never divisible by 17. II. n 2 + 4 is never divisible by 17. Then,
(a) 12 (c) 14
15. In a Mathematics test, the average marks of boys is
x% and the average marks of girls is y% with x ≠ y. If the average marks of all students is z%, the ratio of the number of girls to the total number of students is
z−x y−x z+ y (c) y−x
and X > Y > Z and X > Y > Z and X < Y < Z
(b)
y x x x
PHYSICS 16. Particles used in the Rutherford’s scattering experiment to deduce the structure of atoms (a) had atomic number 2 and were fully ionised (b) had atomic number 2 and were neutral (c) had atomic number 4 and were fully ionised (d) had atomic number 4 and were neutral
17. The number of completely filled shells for the element 16S32 is
(a) 1 (c) 3
(b) 2 (d) 4
18. In an experiment on simple pendulum to determine the acceleration due to gravity, a student measures the length of the thread as 63.2 cm and diameter of the pendulum bob as 2.256 cm. The student should take the length of the pendulum to be (a) 64.328 cm (c) 65.456 cm
(b) 64.3 cm (d) 65.5 cm
19. A uniform metallic wire of length L is mounted in two configurations. In configuration 1 (triangle), it is an equilateral triangle and a voltage V is applied to corners A and B. In configuration 2 (circle), it is bent in the form of a circle and the potential V is applied at diametrically opposite points P and Q. The ratio of the power dissipated in configuration 1 to configuration 2 is A
(d) 36
pentagon, regular hexagon and regular heptagon which are inscribed in a circle of radius 1. Then, and X < Y < Z
z− y− z+ (d) y−
(a)
P
13. Let X ,Y , Z be respectively the areas of a regular
X Y Z (a) < < 5 6 7 X Y Z (b) < < 5 6 7 X Y Z (c) > > 5 6 7 X Y Z (d) > > 5 6 7
(b) 13 (d) 15
Q
B
(a) 2/3
(b) 9/8
(c) 5/4
(d) 7/8
20. Six objects are placed at the vertices of a regular hexagon. The geometric centre of the hexagon is at the origin with objects 1 and 4 on the Xaxis (see figure). The mass of the kth object is mk = ki M cos θ k , where i is an integer, M is a constant with dimension of mass and θ k is the angular position of the k th vertex measured from the positive Xaxis in the counterclockwise sense.
KVPY Question Paper 2017 Stream : SA If the net gravitational force on a body at the centroid vanishes, the value of i is 3
2 y x
4
5
(a) 0
(b) 1
1
6
(c) 2
(d) 3
21. A mirror is placed at an angle of 30° with respect to Y axis (see figure). A light ray travelling in the negative ydirection strikes the mirror. The direction of the reflected ray is given by the vector y
24. Two students P and Q perform an experiment to verify Ohm’s law for a conductor with resistance R. They use a current source and a voltmeter with least counts of 0.1 mA and 0.1 mV, respectively. The plots of the variation of voltage drop V across R with current I for both are shown below. 12 10 8 6 4 2 0 –2 –4 –6 –8 –10 –12 –6
P
V(mV)
18
–4
–2
0
2
4
6
0
2
4
6
I(mA) 30°
(b) $i − 3$j (d) $i − 2 $j
(a) $i (c) 3 $i − $j
22. A total charge q is divided as q1 and q2 which are kept at two of the vertices of an equilateral triangle of side a. The magnitude of the electric field E at the third vertex of the triangle is to be depicted schematically as a function of x = q1 / q. Choose the correct figure.
0.0
0.5 x
0.0
1.0
(c)
0.5 x
1.0
–2
I(mA)
0.5 x
1.0
25. A cylindrical vessel of base radius R and height H has a narrow neck of height h and radius r at one end (see figure). The vessel is filled with water (density ρw ) and its neck is filled with immiscible oil (density ρ0). Then, the pressure at
(d) E
–4
(a) P has only random error(s) (b) Q has only systematic error(s) (c) Q has both random and systematic errors (d) P has both random and systematic errors
E
E
Q
The statement which is most likely to be correct?
(b)
(a)
12 10 8 6 4 2 0 –2 –4 –6 –8 –10 –12 –6
V(mV)
x
O
E
2r
0.0
0.5 x
1.0
0.0
h
23. The refractive index of water in a biology laboratory
tank varies as 133 . + 0002 . / λ2, where λ is the wavelength of light. Small pieces of organic matter of different colours are seen at the bottom of the tank using a travelling microscope. Then, the image of the organic matter appears (a) deeper for the violet pieces than the green ones (b) shallower for the blue pieces than the orange ones (c) at the same depth for both the blue and orange pieces (d) deeper for the green pieces than the red ones
H N
M 2R
(a) M is g (hρ0 + Hρw ) (c) M is gHρw
(b) N is g (hρ0 + Hρw ) (d) N is g
r2
R2 ρw HR + ρ0 hr 2 2
R2 + r 2
19
KVPY Question Paper 2017 Stream : SA 26. Two cars S1 and S2 are moving in coplanar concentric circular tracks in the opposite sense with the periods of revolution 3 min and 24 min, respectively. At time t = 0, the cars are farthest apart. Then, the two cars will be (a) closest to each other at t = 12 min and farthest at t = 18 min (b) closest to each other at t = 3 min and farthest at t = 24 min (c) closest to each other at t = 6 min and farthest at t = 12 min (d) closest to each other at t = 12 min and farthest at t = 24 min
27. In the circuit shown below, a student performing Ohm’s law experiment accidently puts the voltmeter and the ammeter as shown in the circuit below. The reading in the voltmeter will be close to 6V
V
2kΩ
(b) 4.8 V
(a) Only statement III is correct (b) Only statement II is correct (c) Only statements III and IV are correct (d) All statements are correct
8kΩ
CHEMISTRY A
(a) 0 V
Ignoring magnification effects, consider the following statements. (I) First image has been viewed from the planar side of a planoconcave lens and second image from the planar side of a planoconvex lens. (II) First image has been viewed from the concave side of a planoconcave lens and second image from the convex side of a planoconvex lens. (III) First image has been viewed from the concave side of a planoconcave lens and second image from the planar side of a planoconvex lens. (IV) First image has been viewed from the planar side of a planoconcave lens and second image from the convex side of a planoconvex lens. Which of the above statements are correct?
(c) 6.0 V
(d) 1.2 V
31. The IUPAC name for the following compound is
28. The Bhagirathi and the Alaknanda merge at Deoprayag to form the Ganga with their speeds in the ratio 1 : 1: 5. The crosssectional areas of the Bhagirathi, the Alaknanda and the Ganga are in the ratio 1 : 2 : 3. Assuming streamline flow, the ratio of the speed of Ganga to that of the Alaknanda is (a) 7 : 9
(b) 4 : 3
(c) 8 : 9
(d) 5 : 3
29. A long cylindrical pipe of radius 20 cm is closed at its upper end and has an airtight piston of negligible mass as shown. When a 50 kg mass is attached to the other end of piston, it moves down by a distance ∆l before coming to equilibrium. Assuming air to be an ideal gas, ∆l / l (see figure) is close to (g = 10 m/ s2, atmospheric pressure is 105 Pa),
(a) 4,6dimethylheptane (c) 2,4dimethylheptane
32. The stability of carbocations +
(CH3)3C I
∆l
+
+
follows the order (a) III < IV < II < I (c) IV < III < II < I
(b) III < IV < I < II (d) IV < III < I < II
33. The acidity of compounds IIV in water I. ethanol III. phenol follows the order
II. acetic acid IV. acetonitrile (b) I < II < III < IV (d) IV < III < I < II
34. In the following reaction, O
(a) 0.01
(b) 0.02
(c) 0.04
through different lenses such that board is at a distance beyond the focal length of the lens.
KVPY
KVPY
NH2
(d) 0.09
30. The word KVPY is written on a board and viewed
Second image
Br2 KOH
the major product is Br (a)
CO2H (b)
Br First image
+
(CH3)2C(OCH3) CH3CH2CH2CH2 CH3CHCH2CH3 IV III II
(a) IV < I < III < II (c) IV < I < II < III
l
(b) 1,3,5trimethylhexane (d) 2,4,6trimethylhexane
KVPY Question Paper 2017 Stream : SA
20 NH2
CONH2
(c)
H4 P2O6 , respectively are
(d)
(a) +5, +4, +4 (c) +4, +4, +5
Br
Fehling’s test for aldehydes (RCHO) is due to the formation of (b) Cu 2O (d) ( RCOO)2 Cu
36. The reducing ability of the metals K, Au, Zn and Pb follows the order (a) K > Pb > Au > Zn (c) Zn > Au > K > Pb
(b) Pb > K > Zn > Au (d) K > Zn > Pb > Au
37. White phosphorus catches fire in air to produce dense white fumes. This is due to the formation of (a) P4O10 (c) H 3 PO3
(b) PH 3 (d) H 3 PO2
in the shell with the principal quantum number n = 4 is (b) 26
(c) 18
(d) 32
39. At a constant pressure p, the plot of volume (V ) as a function of temperature (T ) for 2 moles of an ideal gas gives a straight line with a slope 0.328 LK −1. The value of p (in atm) is closest to [Gas constant, R = 00821 L atm mol −1 K −1 ] . (a) 0.25 (c) 1.0
(b) 0.5 (d) 2.0
out by using HI as a reducing agent, under acidic conditions? [Given : I2 (s) → 2I− ; E ° = 054 . V] (i) Cu + → Cu (s); E ° = 052 . V (ii) Cr3 + → Cr 2+ ; E ° = − 041 . V (iii) Fe
2+
°
→ Fe ; E = 0.77 V
(iv) Fe 2+ → Fe (s); E ° = − 044 . V (a) (i) and (iii) (c) Only (iii)
(b) (ii) and (iv) (d) Only (ii)
41. C 60 emerging from a source at a speed (v) has a de Broglie wavelength of 11.0 Å. The value of v (in ms−1 ) is closest to [Planck’s constant h = 6626 . × 10 (a) 0.5
(b) 2.5
(c) 5.0
−34
Js] (d) 30
42. The lattice energies of NaCl, NaF, KCl and RbCl follow the order (a) KCl < RbCl < NaCl < NaF (b) NaF < NaCl < KCl < RbCl (c) RbCl < KCl < NaCl < NaF (d) NaCl < RbCl < NaF < KCl
(a) 1.2
(b) 2.4
(c) 4.8
(d) 0.6
45. 1.25 g of a metal (M) reacts with oxygen completely to produce 1.68 g of metal oxide. The empirical formula of the metal oxide is [molar mass of M and O are 69.7 g mol −1 and 16.0 g mol −1 , respectively] (a) M2O (b) M2O3 (c) MO2 (d) M3 O4
46. According to WatsonCrick model, hydrogen bonding in a doublestranded DNA occurs between (a) adenine and guanine (c) cytosine and adenine
(b) adenine and thymine (d) guanine and thymine
47. Which one of the following statements about mitosis is correct? (a) One nucleus gives rise to 4 nuclei (b) Homologous chromosomes synapse during anaphase (c) The centromeres separate at the onset of anaphase (d) Nonsister chromatids recombine
48. Gaseous exchange of oxygen and carbon dioxide
40. Which of the following transformations can be carried
3+
neutralised by y mL of 1M NaOH. The same volume (y mL) of 1M NaOH is required to neutralise 10 mL of 0.6 M of H2 SO4 completely. The normality (N) of the acid X is
BIOLOGY
38. The maximum number of electrons that can be filled (a) 64
(b) +5, +5, +4 (d) +3, +4, +5
44. A solution (5 mL) of an acid X is completely
35. The reddish brown precipitate formed in the (a) Cu (c) CuO
43. The oxidation states of P atom in POCl 3 , H2 PO3 and
between alveolar air and capillaries takes place by (a) active transport (b) diffusion (c) carriermediated transport (d) imbibition
49. Of the periods listed below, which one is the earliest period when ostracoderms, the jawless and finless fishes, appeared? (a) Devonian period (c) Carboniferous period
(b) Cambrian period (d) Silurian period
50. Scurvy is caused by the deficiency of (a) nicotinic acid (c) pantothenic acid
(b) ascorbic acid (d) retinoic acid
51. Optical activity of DNA is due to its (a) bases (c) phosphates
(b) sugars (d) hydrogen bonds
52. The monarch butterfly avoids predators such as birds by (a) changing colour frequently (b) flying away from the predator swiftly (c) producing a chemical obnoxious to the predator (d) producing ultrasonic waves
21
KVPY Question Paper 2017 Stream : SA 53. Filariasis is caused by
58. Match the diseases given in Column I with the
(a) Entamoeba histolytica (b) Plasmodium falciparum (c) Trypanosoma brucei (d) Wuchereria bancrofti
principal causal organisms in Column II and choose the correct combination.
54. Which one of the following conversions does not happen under anaerobic conditions? (a) Glucose to ethanol by Saccharomyces (b) Lactose to lactic acid by Lactobacillus (c) Glucose to CO2 and H 2O by Saccharomyces (d) Cellulose to glucose by Cellulomonas
55. An amount of 18 g glucose corresponds to (a) 1.8 mole (b) 1 mole
(c) 0.18 mole (d) 0.1 mole
56. The number of electrons required to reduce one molecule of oxygen to water during mitochondrial oxidation is (a) 4
(b) 3
(c) 2
(d) 1
57. Which one of the following molecules is derived from
Column I (P) (Q) (R) (S)
Column II
AIDS Syphilis Viral hepatitis Gonorrhoea
(a) Piv, Qiii, Ri, Sii (c) Pi, Qii, Riv, Siii
(i) (ii) (iii) (iv)
HBV Neisseria sp. Treponema sp. HIV
(b) Piv, Qii, Ri, Siii (d) Pi, Qiv, Rii, Siii
59. Chromosomes are classified based on the position of centromere. A chromosome having a terminal centromere is called (a) metacentric (c) submetacentric
(b) telocentric (d) acrocentric
60. Which one of the following options lists the primary
pantothenic acid?
energy source(s) for all forms of life on the earth?
(a) Thiamine pyrophosphate (b) Nicotinamide adenine dinucleotide phosphate (c) Flavin adenine dinucleotide phosphate (d) AcetylCoA
(a) Light, inorganic substances (b) Inorganic substances, organic substances (c) Light, organic substances (d) N2 , CO2
PARTII (2 Marks Questions) MATHEMATICS
64. Let C1, C 2 be two circles touching each other
61. Let ABCD be a trapezium with parallel sides AB and CD such that the circle S with AB as its diameter touches CD. Further, the circle S passes through the midpoints of the diagonals AC and BD of the trapezium. The smallest angle of the trapezium is π 3 π (c) 5 (a)
π 4 π (d) 6 (b)
a c b d radius 1 centred at (0, 0) where a and b are relatively prime integers, c and d are relatively prime integers (that is HCF (a , b) = HCF (c, d) = 1), and the integers b and d are even. Then, the set S
62. Let S be the set of all points , on the circle with
(a) is empty (b) has four elements (c) has eight elements (d) is infinite
30 3 30 , 5 10 6 6 (c) , 2 2 (a)
plane such that the distance between their centers is 2 3 . The area of the region common to both circles lies between (b) 0.65 and 0.7 (d) 0.8 and 0.9
5 7 5 , 2 10 10 17 10 (d) , 3 30 (b)
65. Let a , b, c, d be real numbers between − 5 and 5 such that  a = 4 − 5 − a ,b = 4 + 5 − b ,c = 4 − 5 + c , d  = 4 + 5 + d Then, the product abcd is (a) 11 (c) 121
63. Suppose we have two circles of radius 2 each in the
(a) 0.5 and 0.6 (c) 0.7 and 0.75
externally at the point A and let AB be the diameter of circle C1. Draw a secant BA3 to circle C 2, intersecting circle C1at a point A1 (≠ A), and circle C 2 at points A2 and A3 . If BA1 = 2, BA2 = 3 and BA3 = 4, then the radii of circles C1 and C 2 are respectively
(b) − 11 (d) − 121
PHYSICS 66. Persons A and B are standing on the opposite sides of a 3.5 m wide water stream which they wish to cross. Each one of them has a rigid wooden plank whose mass can be neglected. However, each plank is only slightly longer than 3 m. So, they decide to arrange them together as shown in the figure schematically.
KVPY Question Paper 2017 Stream : SA
22 With B (mass 17 kg) standing, the maximum mass of A, who can walk over the plank is close to A
3m B
given charge Q, so that balls move away from each other with each thread making an angle of 45° from the vertical. The value of Q is close to 1 = 9 × 109 in SI units 4πε 0 (a) 1 µC
(a) 17 kg
(c) 2 µC
(d) 2.5 µC
70. Two parallel discs are connected by a rigid rod of
3.5 m
(b) 65 kg
(b) 15 . µC
(c) 80 kg
(d) 105 kg
67. Two different liquids of same mass are kept in two identical vessels, which are placed in a freezer that extracts heat from them at the same rate causing each liquid to transform into a solid. The schematic figure below shows that temperature T versus time t plot for the two materials. We denote the specific heat of materials in the liquid (solid) states to be C L1 (C S1) and C L 2 (C S 2 ), respectively.
length L = 05 . m centrally. Each disc has a slit oppositely placed as shown in the figure. A beam of neutral atoms are incident on one of the discs axially at different velocities v, while the system is rotated at angular speed of 600 rev/second, so that atoms only with a specific velocity emerge at the other end. Calculate the two largest speeds (in metre/second) of the atoms that will emerge at the other end.
T
v
ω
1
(a) 75, 25
2
(b) 100, 50
(c) 300, 100 (d) 600, 200
t
Choose the correct option given below.
CHEMISTRY
(a) CL1 < CL 2 and CS1 < CS 2 (b) CL1 > CL 2 and CS1 < CS 2 (c) CL1 > CL 2 and CS1 > CS 2 (d) CL1 < CL 2 and CS1 > CS 2
68. A ray of light originates from inside a glass slab and
is incident on its inner surface at an angle θ as shown below.
71. Among the following compounds, E/Z isomerism is possible for (a) 2methylbut2ene (c) 3methylpent1ene
72. In the reaction,
Glass slab
H 3C
2
(a) A
0 x θ
A.
B.
(b) B
0 x θ C.
(c) C
1. NaNH2, ∆ 2. x 3. y
H 3C
CH3
73. Among the following molecules, the one with the
In this experiment, the location x of the spot where the ray hits the screen is recorded. Which of the following correctly shows the plot of variation of x with the angle θ ?
θ
H
(a) x = CH 3OH; y = Pd / BaSO4 , quinoline, H 2 (b) x = CH 3 I; y = Pd / BaSO4 , quinoline, H 2 (c) x = CH 3 I; y = Na in liq. NH 3 (d) x = CH 3OH; y = Na in liq. NH 3
Screen
0 x
C
x and y, respectively are
0 –2
0 x
C
x
θ
(b) 2methylbut1ene (d) 3methylpent2ene
θ D.
(d) D
69. Four identical pendulums are made by attaching a small ball of mass 100 g on a 20 cm long thread and suspended from the same point. Now, each ball is
largest bond angle at the central atom is (a) ClF 3
(b) POCl 3
(c) BCl 3
(d) SO3
74. A compound has the following composition by weight : Na = 18.60%, S = 25.80%, H = 4.02% and O = 51.58%. Assuming that all the hydrogen atoms in the compound are part of water of crystallisation, the correct molecular formula of the compound is (a) Na 2 S2O3 ⋅ 3H 2O (c) Na 2 SO4 ⋅ 10H 2O
(b) Na 2 SO4 ⋅ 5H 2O (d) Na 2 S2O3 ⋅ 5H 2O
75. X g of ice at 0°C is added to 340 g of water at 20°C. The final temperature of the resultant mixture is 5°C. The value of X (in g) is closest to [Heat of fusion of ice = 333 J/g; specific heat of water = 4.184 J/g.K] (a) 80.4
(b) 52.8
(c) 120.6
(d) 60.3
23
KVPY Question Paper 2017 Stream : SA
77. A 25,000 Da protein contains a single binding site for
BIOLOGY
a molecule (ligand), whose molecular weight is 2,500 Da. Assuming high affinity and physiologically irreversible binding, the amount of the ligand required to occupy all the binding sites in 10 mg protein will be
76. Considering ABO blood grouping system in humans, during blood transfusion some combinations of blood groups are compatible (3), whereas the others are incompatible (7). Which one of the following options is correct?
Donor
Recipient
(a)
O O ✗ P A P B AB P
A ✗ ✗ P P
B ✗ P ✗ P
(a) 0.1 mg (c) 10 mg
78. In an in vitro translation experiment, poly (UC) RNA
AB P
template produced poly (SerLeu), while poly (AG) RNA template produced poly (ArgGlu) polypeptide. Which one of the following options represents correct interpretations of the codons assignments for Ser, Leu, Arg and Glu?
✗ ✗ P
(a) SerUCU, LeuCUC, ArgAGA, GluGAG (b) SerCUC, LeuGAG, ArgUCU, GluAGA (c) SerAGA, LeuUCU, ArgGAG, GluCUC (d) SerGAG, LeuAGA, ArgCUC, GluUCU
(b)
Donor
Recipient O A B AB
O
A
B
✗ P P P
✗
✗ P
✗ P P
✗ P
AB ✗ ✗ ✗ ✗
79. A single bacterium is actively growing in a medium that supports its growth to a number of 100 million. Assuming the division time of the bacterium as 3 hours and the lifespan of nondividing bacteria as 5 hours, which one of the following represents the maximum number of bacteria that would be present at the end of 15 hours?
Donor
Recipient
(c)
O A B AB
O P P P P
A
B
AB
✗ P ✗ P
✗
✗
✗ P P
✗ ✗ P
(a) 10
Donor
(d)
A P P ✗ ✗
B P ✗ P ✗
(b) 64
(c) 24
(d) 32
80. A couple has two sons and two daughters. Only one son is colourblind and the rest of the siblings are normal. Assuming colourblindness is sexlinked, which one of the following would be the phenotype of the parents?
Recipient O O P ✗ A ✗ B AB ✗
(b) 1 mg (d) 100 mg
AB P P P P
(a) Mother would be colourblind, father would be normal (b) Father would be colourblind, mother would be normal (c) Both the parents would be normal (d) Both the parents would be colourblind
Answers PARTI 1
(c)
2
(b)
3
(c)
4
(a)
5
(d)
6
(d)
7
(b)
8
(d)
9
(c)
10
(d)
11
(c)
12
(b)
13
(d)
14
(c)
15
(a)
16
(a)
17
(b)
18
(b)
19
(b)
20
(a)
21
(c)
22
(c)
23
(b)
24
(d)
25
(a)
26
(d)
27
(c)
28
(c)
29
(c)
30
(d)
31
(c)
32
(b)
33
(a)
34
(c)
35
(b)
36
(d)
37
(a)
38
(d)
39
(b)
40
(c)
41
(*)
42
(c)
43
(a)
44
(b)
45
(b)
46
(b)
47
(c)
48
(b)
49
(b)
50
(b)
51
(b)
52
(c)
53
(d)
54
(c)
55
(d)
56
(a)
57
(d)
58
(a)
59
(b)
60
(a)
PARTII 61
(d)
62
(a)
63
(c)
64
(a)
65
(a)
66
(c)
67
(b)
68
(a)
69
(b)
70
(d)
71
(d)
72
(c)
73
(a)
74
(d)
75
(d)
76
(d)
77
(b)
78
(a)
79
(d)
80
(c)
* No option is correct.
KVPY Question Paper 2017 Stream : SA
24
Solutions 1. (c) Let triangle T is PQR and other triangle is ABC. A
P
Q
R
B
C
B
C
P
Q
R A
A can taken position if ∆ABC ~ ∆PQR. We can arrange A , B , C in 3! ways = 6 ways Total position of A can take = 3! × 2 = 12 ways
2. (b) We have, n ∈ {2, 3, 4, 5, 6, …, 200} 1 has terminating decimal of n = 2a × 5b n ∴ n = 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 128, 160, 200 ∴ Total number of n = 18
3. (c) We have, a + b + c = 0 and a 2 + b2 + c2 = 1 Now (3a + 5b − 8c)2 + (− 8a + 3b + 5c)2 + (5a − 8b + 3c)2 2 2 2 = 9a + 25b + 64c − 48ac + 30ab − 80bc + 64a 2 + 9b2 + 25c2 − 80ac − 48ab + 30bc + 25a 2 + 64b2 + 9c2 + 30ac − 8ab − 48bc 2 2 2 = 98(a + b + c ) − 98 (ab + bc + ca ) = 98 (a 2 + b2 + c2 ) (a + b + c)2 − (a 2 + b2 + c2 ) − 98 2 0 − 1 = 98(1) − 98 = 98 + 49 = 147 2
4. (a) ∆ ABC ~ ∆ANM A
M
N
B
∴
P
From Eq. (iii), ⇒ ⇒
13 (AC 2 ) AM 2 + MC 2 = 18 MC 2 MC 2
13 (AM + MC )2 = 18 (AM 2 + MC 2 ) AM =5 MC
5. (d) We have, l1 , l2 , l3 ..., ln be the lengths of the side of arbitrary n sided nondegenerate polygon P and l1 l l ln − 1 ln + 2 + 3 + ... + + = n, n ≥ 4 l2 l3 l4 ln l1 Using AM ≥ GM, we get l1 l2 l3 ln + + + ... + 1/ n l2 l3 l4 l1 l1 l2 ln ≥ × × ... × n l1 l2 l3 l1 l ln + 2 + ... + ≥n ∴ l2 l3 l1 ∴
n ≥ n ⇒n = n
So, AM = GM ∴ l1 = l2 = l3 … = ln ∴ The length of sides of P are equal and P is regular polygon of it is cyclic.
6. (d) Let n 2 + 3 is divisible by 17 ∴ n 2 + 3 = 17K [K ∈ N ] ⇒ n 2 = 17K − 3 ⇒ n 2 = 3 (17m − 1) [Q K = 3 m] 3 (17 m − 1) is a perfect square is not possible. ∴ n 2 + 3 is never divisible by 17. n 2 + 4 put n = 9 (9)2 + 4 = 81 + 4 = 85 is divisible by 17. ∴I is true and II is false.
7. (b) We have,
C
Area of ∆ABC AC 2 = Area of ∆ANM AM 2
∆ABC ~ MPC Area of ∆ABC AC 2 …(ii) = Area of ∆MPC MC 2 From Eqs. (i) and (ii), we get Area of ∆ANM AM 2 = Area of ∆MPC MC 2 Area of ∆ANM + Area of ∆MPC Area of ∆MPC AM 2 + MC 2 = MC 2 Now, Area of ∆ANM + Area of ∆MPC = Area of ∆ABC − Area of BNMP 13 (Area of ∆ABC ) AM 2 + MC 2 = ∴ 18 (Area of ∆ MPC ) MC 2
… (i)
HCF (x, y) = 16 LCM (x, y) = 48000
We know, Product of two number = HCF × LCM ∴ xy = 16 × 48000 xy = 16 × 16 × (31 × 23 × 53 ) As HCF of (x, y) = 16 23 can be selected in 1 ways and 31 and 53 can be selected in (1 + 1) (3 + 1) = 8 ways ∴ Number of ordered pairs = 8
8. (d) Let two digits number ab = 10a + b, b ≠ 0 if b is erased. Then, the resulting number is a. ∴ab is divisible by a if ab is multiple of c. ∴Such number are 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99. ∴Total number are 23. Hence,
K < 25
9. (c) There are 90 days from 1 January to 31 March (Nonleap year) If year 13 leap year, then total number of days = 91(13 weeks) But we have 12 Sunday ∴12 weeks ∴Ist Jan will be Monday as there will be 90 days January 1 to 31 March. ∴15th February will be Thursday. 10. (d) Let three digits number be 100x + 10 y + z. According to problem, 100x + 10 y + z = 100x + 10z + y − 36 ⇒ 9 y − 9z + 36 = 0 …(i) ⇒ y− z + 4= 0 ⇒ 100x + 10 y + z = 100z + 10 y + x + 198 …(ii) ⇒ x− z − 2= 0 Now, (100x + 10 y + z ) − (100 y + 10x + z ) = 90(x − y) = 90(6) [Q from Eqs. (i) and (ii)] = 540 ∴ So, on interchanging for digit at tens place and hundred place, the value of number is decreased by 540.
11. (c) We have, Four triangle having sides are (5, 12, 9), (5, 12, 11), (5, 12, 13), (5, 12, 15) A right triangle has maximum area. ∴ Among these the triangle whose sides (5, 12, 13) form a right angled triangle. ∴ It has maximum area.
25
KVPY Question Paper 2017 Stream : SA 12. (b) Let the number of boys and girls in classroom is x and y, respectively. x − x/5 2 4x 2 Given, = = ⇒ y 3 5y 3 x 5 …(i) ⇒ = y 6 x − x/5 5 4x 5 Also, = = ⇒ 2 5 ( y − 44) 2 y − 44 … (ii) ⇒ 8x = 25 y − 1100 From Eqs. (i) and (ii), we get x = 50, y = 60 Let z number of boy leaves so number of boys and number of girls are equal. ∴ 50 − 10 − z = 60 − 44 z = 40 − 16 = 24
13. (d) We have, X , Y , Z be respectively the area of a regular pentagon, regular hexagon and regular heptagon which are inscribed in radius of unit circle.
Now, given
Bx + Gy =z B+G B z− y ⇒ B (x − z ) = G (z − y ) = = G x−z x−z 1 1 G Now, = = = B + G B + 1 z − y+1 x − y G x−z z−x G = ⇒ B+G y−x scattering experiment (GeigerMarsden experiment) are αparticles derived from a tube of radium emanation (or radon). αparticles are helium nuclei 42 He, they are fully ionised and have atomic number 2.
17. (b) Atomic number of 16 S32 is 16. Its electronic configuration using 2n 2 rule is s2 , 2s2 p 6 , 3s2 p 4 16 S = 1 14243 123 Unfilled
So, number of fully filled orbits or shells is 2.
2π 5 1
2π 1 × 1 × sin 5 2 5 2π X = sin 2 5 Similarly, 6 2π and Z = Y = sin 2 6 X 1 2π Y 1 2π Z = sin , = sin , 5 2 5 6 2 3 7 2π 2π 2π sin > sin > sin 5 6 7 X Y Z ∴ > > and X < Y < Z 5 6 7 ∴
18. (b) Length of pendulum = Length of
thread + Radius of bob 2.256 = 63.2 + 1128 . = 63.2 + 2 = 64.328 cm But now the student must apply rule for taking significant digits in a measurement.
X = 5×
7 2π sin 2 7 1 2π = sin 2 7
So, length of pendulum = 64.3 cm.
19. (b) Let a = side length of equilateral
n −1
⇒
C5 + n − 1C6 < nC7 n C6 < nC7 [Q nCr − 1 + nCr = n! n! < (n − 6)! 6! (n − 7)! 7!
In addition or subtraction, Number of digits after decimal in result = Least number of digits after decimal in quantities added
triangle, r = radius of circle and x = resistance per unit length of wire used. L L Then, L = 3a = 2πr or a = and r = 3 2π Now, in case I,
14. (c) Given,
n+1
⇒ n − 6> 7 ⇒ n > 13 ∴ Least value of x = 14
15. (a) Let the number of boy = B and number of girls = G Sum of marks obtained by boys = Bx ∴Sum of marks obtained by girls = Gy
RAB =
A
Cr ] – V
+
B
Equivalent resistance across AB is ax × 2ax RAB = (ax 2ax) = ax + 2ax 2a 2x2 2 = = ax 3ax 3
...(i)
In case II, P
Q
+
16. (a) Particles used in Rutherford’s
[Ne]
1
2 L × ×x 3 3 Power dissipated is 9V 2 V2 P1 = = RAB 2Lx
⇒
–
V
πrx × πrx π 2r 2x2 = πrx + πrx 2 πrx L 1 1 Lx x= = πrx = π × 2 2 2π 4 So, power dissipated is 4V 2 V2 P2 = = RPQ Lx
RPQ = ( πrx πrx) =
Ratio of power dissipated in two cases is P1 9V 2 / 2Lx 9 = = 8 P2 4V 2 / Lx
20. (a) For a mass m at centroid of hexagon (at origin), net force is zero when ΣFx = 0 and ΣFy = 0. 3 2 r 4
5
m
mk=ki⋅M⋅ cos θk
O
1
6
Now, ΣFx = sum of all xcomponents of forces on m due to masses at vertices of hexagon. Gm = 2 (Σ (ki M cos θk  ⋅ cos θk )) r GmM i (1 cos 0°⋅ cos 0° + 2 i cos 60°  = r2 ⋅ cos 60°+ 3 i cos 120° ⋅ cos 120° + 4 i cos 180° ⋅ cos 180° + 5 i cos 240° ⋅ cos 240° + 6 i cos 300° ⋅ cos 300° ) i 3i 5i 6i GMm i 2 = ⋅ 1 + − − 4i − + 2 4 4 4 4 r As ΣFx = 0, for net force on m to be zero. we have 5i 6i 2i 3i − − 4i − + =0 1i + 4 4 4 4 Above equation is satisfied with i = 0.
KVPY Question Paper 2017 Stream : SA
26 21. (c) Following laws of reflection, reflected ray makes an angle of 30° with mirror as shown below.
Now, µ orange < µ blue ⇒ (Apparent depth)blue < (Apparent depth)orange
24. (d) For P errors are both positive and
A 30°
Incident ray
N
60° x
O
30° 30 °
Reflected ray B
So, if a vector x $i + y$j is along the reflected ray, then y 1 y tan (−30° ) = or = − x 3 x This is correct with option (c). 22. (c) When q1 and q2 are the magnitudes of charges at two vertices of an equilateral triangle of side a, magnitude of electric field at third vertex is E2
negative. For Q errors are only positive. So, P has both random and systematic errors. 25. (a) Pressure is same at all the points of base. i.e. Pressure at M = Pressure at N Also, pressure applied anywhere to the fluid is equally transmitted in all directions. So, pressure at base = pressure due to oil column of height h + pressure due to water column of height H. ⇒ ρo gh + ρw gH ⇒ g(ρ0⋅ h + ρw H )
26. (d)
T=24 min S2
27. (c) The resistance of voltmeter is very high and resistance of ammeter is very low. When ammeter is put in parallel to 8 kΩ resistor, nearly whole of current goes through the ammeter. Hence, circuit is equivalent to following. Low resistance V High resistance
So, maximum potential drop occurs in the voltmeter (high resistance). Hence, reading of voltmeter is nearly 6 V.
28. (c) As water is not stored anywhere. So, volume flow rate of Ganga = volume flow rate of Bhagirathi + volume flow rate of Alaknanda Bha
E E 1
S1
q1 2
q2 2
2
kq1 + kq2 + 2k q1 q2 cos 60° a a a2 k 2 2 q1 + q2 + q1 q2 = a k q12 + (q − q1 )2 + q1 (q − q1 ) = a 2 q k q1 = + 1− 1 q q aq 2 3 k q1 1 = − + 4 aq q 2 q 1 So, field is minimum when 1 = . q 2 E=
This condition is satisfied in graph (c).
At t = 12 min, car S1 has completed three rounds and it is at its position. At t = 12 min, car S2 completed half round and it is at diametrically opposite point as shown below.
S1
S2 Positions of cars at t = 12 min
So, cars are closest at t = 12 min. At t = 24 min, cars S1 and S2 are both at their initial positions and so are farthest, as shown below. S2
23. (b) As, refractive index, µ = 133 . +
0.002
λ2 So, µ is more for small wavelengths. i.e. µ orange < µ green < µ blue real depth apparent depth 1 ⇒ Apparent depth ∝ µ
As,
S1
µ=
Hence, cars are farthest from each other at t = 24 min.
girat
hi
da
nan Alak
T=3 min
Positions of cars at t = 0s
2kΩ
+ –
Ab
vb
Aa
va
Ganga
Ag
vg
∴By equation of continuity, we have ...(i) ⇒ Ag vg = Ab vb + Aa va It is given that area of flow of Ganga, Alaknanda and Bhagirathi are in ratio, Ag : Aa : Ab = 3 : 2 : 1 or
Ag = 3x, Aa = 2x, Ab = x
Also, ratio of speeds of Bhagirathi and Alaknanda is 3 vb : va = 1 : 2 3 or vb = y, va = y 2 Substituting these values in Eq. (i), we get 3 3x ⋅ vg = x ⋅ y + 2x ⋅ y = 4xy 2 4 So, vg = y 3 ∴ Ratio of speed of Ganga to that of Alaknanda is 4 y vg 8 = 3 = 3 va y 9 2
29. (c) Initially pressure inside the cylinder is atmospheric pressure p0 . When mass m is attached to piston and it comes down by a distance ∆l, let pressure is p.
27
KVPY Question Paper 2017 Stream : SA Then, in equilibrium, p0V 0 = pV ⇒ p0 (A ) (l) = pA (l + ∆l) l
p0
∆l Initially m Finally
So, final pressure will be p0 Al p0 l = p= A (l + ∆ l) (l + ∆ l) In equilibrium, weight of mass m is balanced by force of suction due to reduced pressure p. Q ( p0 − p ) A = mg p0 l p A l +1 ⇒ p0 − A = mg ⇒ 0 = l + ∆l ∆l mg 105 × π × (20 × 10−2 )2 l = +1 ∆l 50 × 10 22 × 8 l ⇒ − 1= ∆l 7 169 l ∆l or ⇒ = ≈ 0.04 7 ∆l l ⇒
30. (d) For a planoconcave lens, when view is from concave side. Radius of curvature of surface 1 is R1 = ∞ and radius of curvature of surface 2 is R2 = R. Image
∴First image appears same when viewed from plane or curved side of a planoconcave lens. Similarly, a planoconvex lens is a converging lens from both side view. So, second image appears always inverted in both cases.
31. (c) The IUPAC name for the following compound is 2
4
6 5
7
1
3
2, 4dimethylheptane
32. (b) As the size of alkyl group goes on increasing, the + I effect exerted by it becomes strong and, thus the carbocation will be more stable. So, among 1°, 2° and 3° carbocation, 3° will be most stable. stable among the given carbocations due to resonance stabilisation. (CH3)2
+
C
O (CH3)2
CH3 C
+
O
CH3
Thus, the correct order of stability of carbocations will be +
+
CH3CH2CH2CH2 < CH3CHCH2CH3 < (2°) IV +
So, focal length of lens using 1 1 1 3 = (µ − 1) − µ = for glass 2 f R1 R2 we have, when viewed from curved side, −R f = (µ − 1) R1 = − R , R2 = ∞.
When viewed from plane side, 1 1 = (µ − 1) − R f −R f = ⇒ (µ − 1)
NH2 NH2
Image
Br2 KOH
Benzamide
Aniline
This reaction is known as Hofmann bromamide reaction. It is used for preparing amine containing one carbon less than the starting amide. In this reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to Natom.
35. (b) Fehling’s reagent is a mixture of aqueous copper sulphate and alkaline sodium potassium tartarate. When an aldehyde is heated with Fehling’s reagent a reddish brown precipitate is obtained and the aldehydes are oxidised to corresponding carboxylate anion. This reddish brown precipitate is due to the formation of copper oxide. RCHO + 2Cu2++5OH–
(3°) I
II
33. (a) The acidity of compounds in water depends upon the ease with which it can lose H + ions. Acetic acid is the strongest acid as the negative charge on carboxylate ion (conjugate base) is delocalised over two oxygen atoms. Hence, H+ ion can be easily lost. The next strongest acidic compound phenol. This is because the phenoxide ion is resonance stabilised. This easily allows the H to leave as H+ ion. Among acetonitrile and ethanol, ethanol is more acidic, this is because in ethanol the Hatom is directly attached to more electronegative atom, O. Thus, the correct order of acidity of compound IIV in water will be IV < I < III < II.
∆
RCOO–
(Fehling's solutions)
+
(CH3)3C+ < (CH3)2C(OCH3)
Object
O
+ Cu2O↓ +3H2O
Although (CH 3 )2 C (OCH 3 ) will be highly
(1°) III
and
34. (c)
Hence, image appears erect in both cases.
p
Object
So, a planoconcave lens acts like a diverging lens weather object is viewed from plane side or curved side.
Red brown (ppt).
36. (d) The reducing ability of metals can be determined by electrochemical series. In this series, various elements are arranged according to their decreasing values of standard reduction potentials. The reducing ability of the metal increases as you go up the series. The increasing order of E° values of given metals are, K < Zn < Pb < Au. Thus, the correct order of reducing ability of metals K, Au, Zn and Pb follows the order K > Zn > Pb > Au.
37. (a) White phosphorus is highly reactive and catches fire when exposed to air and produces white dense fumes of phosphorus oxide, P4 O10 . P4 + 5O2 → P4 O10
38. (d) The maximum number of electrons that can be filled in the shell with principle quantum number, n = 2n 2 For n=4 Maximum number of electrons = 2(4)2 = 32 39. (b) According to ideal gas equation pV = nRT ⇒
V nR = = slope T p
Given, slope = 0.328, n = 2 2 × 0.0821 nR p= = = 0.500 atm slope 0.328
∴
KVPY Question Paper 2017 Stream : SA
28 40. (c) The more positive E° value of metal, feasible transformation can be carried out by using HI as reducing agent under acidic conditions. As E° = 0.77 V (Fe3 + → Fe2+ ) is more positive than, E° = 0.54 (I2 (s) → 2I− ) thus can be used for carrying out transformation as it is the strongest oxidising agent among the other given options. 41. (*) According to deBroglie wavelength h mv = 12 × 60 = 720 g λ=
Mass of C 60
o
Given, λ = 110 . A = 11 × 10−10 m h = 6.626 × 10−34 Js = 6.626 × 10−34 kg m 2s −1 6.626 × 10−34 kgm 2s −1 h ∴ v= = mλ 720 × 10−3 kg × 11 × 10−10 m = 0.8 × 10−18 m/s No option is correct in the given format as the value of wavelength is given in Å which gives the large difference in answer.
42. (c) Lattice energy is the energy required to completely separate one mole of a solid ionic compound into gaseous constituent. Lattice energy increases with decrease in the size of ions. This is because as the size of ion is less, intermolecular distance will be less and so forces of attraction is greater. Thus, the correct increasing order of lattice energies is, RbCl < KCl < NaCl < NaF.
43. (a) Let the oxidation state of Patom in POCl3 , H2PO3 and H4 P2O6 be x. (i) POCl 3 x + 1 (−2) + 3(−1) = 0 x− 2− 3= 0 x=+5 (ii) H 2 PO3 2(1) + x + 3(−2) = 0 2+ x− 6= 0 x=+4 (iii) H 4 P2O6 4(1) + 2x + 6(−2) = 0 4 + 2x − 12 = 0 2x = 8 x=+4
44. (b) Number of equivalents = M × V × acidity/basicity Number of equivalents of NaOH = 1× y × 1= y
Number of equivalents of H 2 SO4 = 0.6 × 10 × 2 = 12 (Number of equivalents) NaOH = (Number of equivalents) H 2 SO 4 = y = 12 mL Also, number of equivalents of acid = Number of equivalents of NaOH N × 5 = 1 × 12 × 1 [N = M × basicity] 12 N = = 2.4 5
45. (b) M + O2 → MO2 1.25 × 100 = 74.4% . 168 Percentage of oxygen in oxide = 100 − 74.4% = 25.6% To calculate empirical formula Percentage of M =
Eleme % of nt element
M
O
74.4
25.6
Simplest Simplest At mass Moles whole molar of of no. ratio element element
69.7
74.4 697 . = 1.06
1.06 = 1 1× 2= 2 1.06
16
25.6 16 = 1.6
1.6 1.06 = 150 .
1.50 × 2 =3
∴Empirical formula of metal oxide is M2O3 .
46. (b) In 1953, JD Watson and FHC Crick proposed a 3D model of physiological DNA. They proposed that DNA is a doublestranded helical molecule. It consists of two sugarphosphate backbones on the outside, held together by hydrogen bonds between pairs of nitrogenous bases on the inside. The bases adenine (A) always pairs with thymine (T) by two hydrogen bonds and guanine always pairs with cytosine (C) by three hydrogen bonds. This complimentarity is known as the base pairing rule. 47. (c) In anaphase, sister chromatids separate from centromeres so, number of chromosome becomes double. Other statements about mitosis can be corrected as Mitosis is a single nuclear division that results in two nuclei. Synapsis takes place during prophaseI of meiosis not during mitosis. Nonsister chromatids recombine during prophaseI of meiosis. During mitosis, each sister chromatid separates and moves to opposite pole of the cell at anaphase. 48. (b) Gaseous exchange occurs at the
alveoli in the lungs and takes place by diffusion. The alveoli are surrounded by capillaries so, oxygen and carbon dioxide
diffuse between the air in the alveoli and the blood in the capillaries. Diffusion is the movement of gas from an area of high concentration to an area of low concentration.
49. (b) The class Ostracodermi is represented by the fossil vertebrates of late Cambrian period. The earliest known vertebrates to appear in fossil record were jawless primitive fishlike animals collectively called ostracoderms. These animals resembled the present day cyclostomes (lampreys and hagfishes) in many respects. 50. (b) Scurvy is caused by the deficiency of vitaminC (Ascorbic acid) in the body. It can lead to anaemia, debility, exhaustion, spontaneous bleeding, pain in the limbs and especially the legs, swelling in some parts of the body and sometimes ulceration of the gums and loss of teeth.
51. (b) DNA polymer is made up of nitrogenous base, a sugar and one or more phosphate. Optical activity results due to the molecular asymmetry. The nucleic acid bases have a plane of symmetry. Hence, they do not induce optical activity. Sugars are asymmetric and cause optical activity of DNA.
52. (c) The monarch butterfly avoids predators such as birds by producing a chemical obnoxious to the predator. Monarchs lay their eggs on milkweed (swan plants), a member of the genus Asclepias. As the caterpillars eat the milkweed leaves, they ingest chemicals called cardiac glycosides. Birds or other animals that eat the caterpillars (or milkweed itself) become sick and vomit. The caterpillars sequester (hold onto) this toxin as they pupate and the toxins are transferred to the adult butterflies. Birds or other creatures that eat the monarchs become sick, so they learn to leave both the butterflies and larvae alone. 53. (d) Filariasis is caused by Wuchereria bancrofti. It lives in lymphatic vessels and causes swelling of lower limbs and scrotum. Entamoeba histolytica causes amoebiasis. Plasmodium falciparum causes malaria. Trypanosoma brucei causes African sleeping sickness. 54. (c) Conversion of glucose to CO2 and H 2O by Saccharomyces is a reaction which takes place in aerobic conditions, i.e. in the presence of oxygen. C6 H12O6 + 6O2 → 6CO2 + 6H2O
29
KVPY Question Paper 2017 Stream : SA 55. (d) A mole is the quantity of a substance whose weight in grams is equal to the molecular weight of the substance. 1 mole is equal to 1 moles Glucose, or 180.15588 grams. ∴ 18 g of glucose = x mole × 180 g 18 x mole = . mole = 01 180 ∴An amount of 18 g glucose corresponds to 0.1 mole.
56. (a) Four electrons are required to reduce one molecule of oxygen to water during mitochondrial oxidation. O2 + 4e− + 4H+ → 2H2O This process mentioned above takes place during oxidative phosphorylation. It is the metabolic pathway in which cells use enzymes to oxidise nutrients, thereby releasing energy which is used to produce ATP.
contain specific combinations of inorganic elements including carbon, hydrogen, nitrogen and oxygen that combine to form proteins and nucleic acids.
61. (d) Given, ABCD is a trapezium where AB is parallel to CD. A circle S with AB as diametre touch CD and also circle passes through the midpoints of diagonal AC and BD. AR = RC ∠ARB = 90° ∴∆ABC is isosceles … (i) AB = BC
58. (a) HIV is the causative organism for AIDS. Syphilis is a bacterial infection caused by Treponema sp. It spreads by sexual contact that starts as a painless sore. Viral hepatitis caused by HBV is an infection that causes liver inflammation and damage organs. Gonorrhoea is caused by Neisseria sp. It is sexually transmitted bacterial infection that if let untreated may cause infertility.
59. (b) Telocentric chromosome is a chromosome like a straight rod with the centromere in terminal position. Metacentric chromosome is a Xshaped chromosome, with the centromere in the middle so that the two arms of the chromosomes are almost equal. Acrocentric chromosome is a chromosome in which the centromere is located quite near one end of the chromosome. Submetacentric chromosome is a chromosome whose centromere is located near the middle.
60. (a) Living organisms require energy to grow, reproduce and respond to the environment. Energy sources include primarily light and inorganic compounds. The most common source of energy on the earth is photosynthesis, which transforms sunlight into food. Life forms usually
63. (c) Given, Two circle each of radius is 2 and difference between their centre is 2 3 1 AB = 2 3 ⇒ AC = AB 2 AC = 3 P 2 A
O
A
D
P
M
C
a c x2 + y2 = 1, , lie on circle. b d b
+
c2
=1 d2 c 1 2 ⇒ =± b − a2 d b c and d are relatively prime. c ∴ is rational. d So, b2 − a 2 = λ2 [Q b is even;∴a is odd] b is even, a is odd. ∴ λ2 is odd ⇒ b2 = λ2 + a 2 ⇒ b2 = (2k + 1)2 + (2m + 1)2 ⇒ b2 = 4k 2 + 4k + 1 + 4m2 + 4m + 1 ⇒ b2 = 4 (k 2 + m2 + k + m) + 2 ∴b is even;∴b2 is multiple of 4. 2
B
√3
AC 3 = AP 2 θ = 30° Area of common region = 2 (Area of sector − Area of ∆APQ) 1 60 = 2 × π(2)2 − × (2)2 × sin 60° 360 2 cosθ =
In ∆APC,
62. (a) Let the equation of circle is
a2
C
Q
Similarly, in ∆ABD BQ = QD ∠AQB = 90° ∴∆ABD is isosceles. … (ii) ∴ AB = AD From Eqs. (i) and (ii), AB = BC = AD ∴Trapezium is isosceles. In ∆ADM, AM OP OP sin ∠ADM = = = AD AD 2OP Q OP = 1 AB 2 1 ⇒ sin (∠ADM ) = 2 π ∠ADM = 30° = ⇒ 6
∴
2
θ √3
B
R
Q
57. (d) VitaminB5 is pantothenic acid or pantothenate, that is required in the synthesis of acetyl CoA. In all living organisms, CoA is synthesised in a five step process that requires four molecules of ATP, pantothenate and cysteine.
But 4 (k 2 + m2 + k + m) + 2 is not multiple of 4. ∴ Not possible. ∴S is empty set.
4π 4 3 = 2 − 4 6 2 . ) − (173 . ) = 2 (314 3 = 2 (2.09 − 173 . ) = 2 (0.36) = 0.72 ∴Area of region lie between 0.7 and 0.75.
64. (a) Given, AB is diameter of circle C1 .
M B
P
C1
A1
A2
A
N
Q
A3
C
C2
BA1 = 2 BA2 = 3 BA3 = 4 Let radius of circle C1 = r1 and radius of circle C2 = r2 ∴ BA = 2r1 and AC = 2r2 1 BM = BA1 = 1 ⇒ 2 1 BN = BA2 + A2A3 ⇒ 2 1 7 = 3+ = 2 2
KVPY Question Paper 2017 Stream : SA
30 In ∆BMP and ∆ BNQ, ∆BMP ~ ∆BNQ BM BP = ∴ BN BQ r1 1 = ⇒ 7 / 2 2r1 + r2
67. (b) Let Q = rate of heat removal. ⇒
⇒ 2r2 = 3r1 Now, BA2 × BA3 = BA × BC ⇒ 3 × 4 = 2r1 (2r1 + 2r2 ) ⇒ 12 = 4 (r12 + r1 r2 ) 2 ⇒ r1 + r1 r2 = 3 From Eqs. (i) and (ii), we get r1 =
6 = 5
… (i)
Slope of graph 1 in solid state is more
mg
F tan 45° = e mg ⇒
mg = Fe 2kq2 ⇒ mg = 2 x ⇒ mgx2 = 2kq2 l where, x = l sin θ = 2
1
30 3 30 and r2 = 5 10
2
t
So, substituting values, we get
a =
4−
5− a
b =
4+
5− b
From graph, ∴ and
c =
4−
5+ c
68. (a) As θ increases, angle of incidence
d =
4+
5+ d
20 × 10−2 ⇒100 × 10−3 × 10 × 2
CS1 < CS 2 CL1 > CL 2
(i = 90 − θ) decreases. Initially upto i = ic angle of critical incidence, reflection takes place and x is positive. Also, x increases till θ is such that i = ic , after that refraction takes place and x becomes negative.
a 4 − 8a 2 + a + 11 = 0
Similarly, squaring other given equation and solving we can say that When θ is less and TIR occur
x4 − 8x2 + x + 11 = 0 ∴ The product of roots abcd = 11
66. (c) When planks are arranged as given in question, we have following situation. Pivot 17 kg B
m A
θc
x
θ
a , b, − c, − d are roots of equation
When θ is large and TIR clean occur
O
kq2 —2 x kq2 ——2 (√2x)
45° 45° x kq2 —2 x
0.5 m 2.5 m
Let m = maximum mass of A. Then, for safe crossing, mg × 0.5 = 17 × g × 2.5 17 × 2.5 m= = 85 kg ⇒ 0.5 So, a man of mass upto 80 kg can pass over planks.
Fe =
kq2 ( 2x)2 =
+
kq2 2x2
2kq x2 +
1 = + 2
= 2 × 9 × 109 × q2 q2 =
⇒
−2
10
9 × 109
q2 > 10−12 (slightly higher than 10−12 ) ⇒ q > 10−6 C So, nearest answer is 1.5 µC.
⇒
70. (d) Time at which discs gaps are alined, π 3π 5π t= , , ,K ω ω ω So, speeds of atoms that emerges on other side are 0.5 = 600 ms −1 v1 = π × 2π 600 0.5 and = 200 ms −1 v2 = 3π × 2π 600 substituents (x, y) at each end of the C == C, then only they can show E/Z isomerism. (a) 2methylbut2ene x CH3
⋅ cos 45°
H x C
C CH3 y
y CH3
2
2
71. (d) If alkenes have two different
x
x
3.5 m
O x
69. (b) Electrostatic force on any of the ball is (let x = separation between two adjacent balls).
x
Fe
x
Slope of graph 2 in liquid state is more
… (ii)
On squaring, we get a2 = 4 − 5 − a = a2 − 4 = − 5 − a Again squaring, we get a 4 − 8a 2 + 16 = 5 − a
i.e.
45° l
Comparing this with y = mx, 1 Slope of Tt line ∝ Specific heat T
65. (a) Given,
⇒
Q ⋅ t = mcT Q ⋅t T = mc
Then,
kq x2
kq2 2kq2 2 2 ≈ 2 x x
As each ball is at an angle of 45° from each other, so in equilibrium, we have
As x = y, this alkene will not show E/Z isomerism. (b) 2methylbut1ene x H C y H
C
CH2CH3 x CH3 y
31
KVPY Question Paper 2017 Stream : SA Here also x = y ∴ It will not show E/Z isomerism. (c) 3methylpent1ene
76. (d) Blood groupO individuals are called universal donor as they can give blood to person with blood groupA, B, AB and O. Blood groupAB individuals can only give blood to persons with blood groupAB but can receive blood from all other blood groups. Therefore, the correct table for the blood transfusion compatibility for ABO blood group system in human is
H x C
C CH2(CH3)CH2CH3 y
y H
This compound will not show E/Z isomerism. (d) 3methylpent2ene x H C
C
Recipient
CH2CH3 x CH3 y
y H 3C
(d)
As x ≠ y, thus this alkene will show E/Z is isomerism.
72. (c) CH3 — C
C—H
ρ σ
(i) Na NH2
CH3 C
σ ρ C Na δ
HC==CH
(iii) Na/liq. NH3
(y)
CH3C
O A B AB
A P P ✗
O P ✗ ✗ ✗
✗
B P ✗ P ✗
AB P P P P
77. (b) Assuming x as the amount of ligand to occupy all the
(ii) CH3 — I (x) – +
CH3
Donor
x H
δ
CCH3 +NaI
CH3 (Trans form)
Thus, in the given reaction, x = CH3 I and y = Na / liq. NH3.
binding sites in 10 mg protein. x mg Ligand molecular weight = Protein in grams Protein molecular weight 2500 x= × 10 = 1mg 25000
78. (a) Serine is coded by UCU, UCC, UCA, UCG, AGU, AGC
the central atom. More is the electronegativity of the central, larger is the bond angle. Thus, among the given central atom Cl has the highest electronegativity. Therefore, ClF3 has the largest bond angle at the central atom.
Leucine is coded by CUU, CUC, CUA, CUG, UUA, UUG Arginine is coded by AGA, AGG, CGU, CGC, CGA, CGG Glutamic acid is coded by GAA, GAG Therefore, option (a) is the correct interpretation of the assigned amino acids. SerUCU, LeuCUC, ArgAGA, GluGAG
74. (d)
79. (d) The number of bacteria after 15 hours will be 32. This
73. (a) The bond angle depends upon the electronegativity of
Elements
% of element
Na
18.6
S
25.8
O
51.58
H
4.02
At mass Simplest Simplest Moles of of molar whole element element ratio no. 18.6 0.8 1 × 2= 2 = 0.8 =1 23 23 0.8 25.8 0.8 1× 2 = 2 = 0.8 =1 32 32 0.8 5158 . 3.22 = 4 4× 2= 8 16 16 0.8 = 3.22 4.02 4.02 = 5 5 × 2 = 10 1 1 0.8 = 4.02
Thus, the empirical formula of compound is Na 2 S2 H10O8 . As it is given the all the hydrogen atoms in the compound are part of water of crystallisation, therefore molecular formula will be Na 2 S2O3 ⋅ 5H 2O.
75. (d) Given, latent heat of fusion of ice = 333 J/g Specific heat of water = 4184 J/g K . First X g of ice at 0°C melts and then its temperature increases by gaining heat from 340 g of water at 20°C. ∴Energy gained by X g of ice = energy lost by 340 g of water [E = mc ∆T ] X (333) + X × 4184 . (278 − 273) = 340 × 4184 . (293 − 278) 333X + 20.92X = 21338.4 ⇒ 353.92X = 21338.4 X = 60.29 ≈ 60.3 g
happens as each bacterium doubles up after every 3 hours, i.e. it 15 will double = 5 times in 15 hours. 3 Therefore, the sequence in which the growth of bacteria taking place will be 1
3 hours
2
3 hours
4
3 hours
8
3 hours
16
3 hours
32
80. (c) Colourblindness is a Xlinked recessive disease, i.e. an heterozygous mother does not show the disease and is a carrier. But a father cannot be a carrier of the disease as it has single Xchromosome. In the given question, the son is colourblind which means it had inherited X c from the mother. But another son is normal. This shows that the mother is heterozygous for the disease. The expected cross for the question will be ×
XCX
XY
Normal carrier mother
XCX
Normal father
XCY
Normal Colourblind carrier son daughter
XX
XY
Normal Normal daughter son
This shows that both the parents would be normal if they have one colourblind son and one normal son.
KVPY Question Paper 2017 Stream : SA
32
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2017 Stream : SA (Nov 05) MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS 1. A quadrilateral has distinct integer side lengths. If the secondlargest side has length 10, then the maximum possible length of the largest side is (a) 25
(b) 26
(c) 27
(d) 28
200 ! 2. The largest power of 2 that divides is 100 ! (a) 98
(b) 99
(c) 100
(d) 101
3. Let a1 , a 2 , a3 , a 4 be real numbers such that + + + = 1. Then, a1 + a 2 + a3 + a 4 = 0 and the smallest possible value of the expression (a1 − a 2 )2 + (a 2 − a3 )2 + (a3 − a 4 ) + (a 4 − a1 )2 lies in a12
a 22
a32
a 42
the interval (a) (0, 1.5)
(b) (1.5, 2.5) (c) (2.5, 3)
(d) (3, 3.5)
4. Let S be the set of all ordered pairs (x, y) of positive integers satisfying the condition x2 − y2 = 12345678. Then, (a) S is an infinite set
(b) S is the empty set (c) S has exactly one element (d) S is a finite set and has at least two elements.
5. Let A1 A2 A3 ... A9 be a ninesided regular polygon with side length 2 units. The difference between the lengths of the diagonals A1 A5 and A2 A4 equals
(a) 2 +
12 (b) 12 − 2
(c) 6
(d) 2
6. Let a1 , a 2 , ... , a n be n nonzero real numbers, of which p are positive and remaining are negative. The number of ordered pairs ( j, k), j < k, for which a j a k is positive, is 55. Similarly, the number of ordered pairs ( j, k), j < k, for which a j a k is negative, is 50. Then, the value of p2 + (n − p)2 is (a) 629
(b) 325
(c) 125
(d) 221
7. If a , b, c, d are four distinct numbers chosen from the set {1, 2, 3,... ,9}, then the minimum value of (a)
3 8
(b)
1 3
(c)
13 36
(d)
25 72
a c + is b d
33
KVPY Question Paper 2017 Stream : SA 8. If 72x ⋅ 48 y = 6xy , where x and y are nonzero rational numbers, then x + y equals (a) 3
(b)
10 3
(c) − 3
(d) −
10 3
9. Let AB be a line segment of length 2. Construct a semicircle S with AB as diameter. Let C be the midpoint of the arc AB. Construct another semicircle T external to the ∆ ABC with chord AC as diameter. The area of the region inside the semicircle T but outside S is (a)
π 2
(b)
1 2
(c)
π 2
(d)
1 2
10. Let r (x) be the remainder when the polynomial 135
x
+x
125
−x
115
+ x + 1 is divided by x − x. Then, 5
3
(a) r (x) is the zero polynomial (b) r (x) is a nonzero constant (c) degree of r (x) is one (d) degree of r (x) is two
11. It is given that the number 43361 can be written as a product of two distinct prime number p1 , p2. Further, assume that there are 42900 numbers which are less than 43361 and are coprime to it. Then, p1 + p2 is (a) 462
(b) 464
(c) 400
(d) 402
12. Let ABC be a triangle with ∠C = 90° . Draw CD perpendicular to AB. Choose points M and N on sides AC and BC respectively such that DM is parallel to BC and DN is parallel to AC. If DM = 5, DN = 4, then AC and BC are respectively equal to (a)
41 41 , 4 5
(b)
39 39 , 4 5
(c)
38 38 , 4 5
(d)
37 37 , 4 5
13. Let A, G and H be the arithmetic mean, geometric mean and harmonic mean, respectively of two distinct positive real numbers. If α is the smallest of the two roots of the equation A(G − H )x2 + G(H − A) x + H ( A − G) = 0 then, (a) − 2 < α < − 1 (c) − 1 < α < 0
PHYSICS 16. Consider the following statements (X and Y stand for two different elements):
(I)
65 32 X
65 and 33 Y are isotopes.
(II)
86 42 X
and
174 85 X
85 42 Y
and have the same number of neutrons. 235 (IV) 235 92 X and 94 Y are isobars.
(III)
Which of the above statements are correct? (a) Only statements II and IV are correct (b) Only statements I, II and IV are correct (c) Only statements II, III and IV are correct (d) All statements are correct
17. A student performs an experiment to determine the acceleration due to gravity g. The student throws a steel ball up with initial velocity u and measures the height h travelled by it at different times t. The graph the student should plot on a graph paper to readily obtain the value of g is (b) h versus t 2 (d) h / t versus t
(a) h versus t (c) h versus t
18. A person goes from point P to point Q covering 1 / 3 of the distance with speed 10 km/h, the next 1/3 of the distance at 20 km/h and the last 1/3 of the distance at 60 km/h. The average speed of the person is (a) 30 km/h (b) 24 km/h
drawn with centre O on the extended line CD and passing through A. If the diagonal AC is tangent to the circle, then the area of the shaded region is O
B
Q
R
S
Note Letters P , Q, R and S are used only to denote the endpoints of the lines.
D
Q
R
S
6− π (d) 4
P
Q S
(A)
C
7− π (c) 4
P
Which of the following represents schematically the image correctly?
X
8− π (b) 6
(d) 12 km/h
length lines PQ and RS in a convex mirror (see figure).
P A
(c) 18 km/h
19. A person looks at the image of two parallel finite
(b) 0 < α < 1 (d) 1 < α < 2
14. In the figure, ABCD is a unit square. A circle is
9− π (a) 6
are isotopes.
177 88 Y
R (B)
Q
P
P
Q
S
R
R
S
15. The sum of all noninteger roots of the equation x5 − 6x4 + 11x3 − 5x2 − 3x + 2 = 0 is (a) 6
(b) −11
(c) − 5
(C)
(d) 3
(a) A
(b) B
(D)
(c) C
(d) D
KVPY Question Paper 2017 Stream : SA
34
atmospheric pressure, two copper hemispheres were tightly fitted to each other to form a hollow sphere and the air from the sphere was pumped out to create vacuum inside. If the radius of each hemisphere is R and the atmospheric pressure is p, then the minimum force required (when the two hemispheres are pulled apart by the same force) to separate the hemispheres is (c) pπR 2
(d)
p πR 2 2
21. Positive point charges are placed at the vertices of a star shape as shown in the figure. Direction of the electrostatic force on a negative point charge at the centre O of the star is 2q a
O 3q q
(a) towards right (c) towards left
(b) vertically up (d) vertically down
22. A total solar eclipse is observed from the earth. At the same time an observer on the moon view’s the earth. She is most likely to see (E denotes the earth) E (A)
E
(C)
(a) A
(I) First image has been viewed from the planar side of a planoconvex lens and second image from the convex side of a planoconvex lens. (II) First image has been viewed from the concave side of a planoconcave lens and second image from the convex side of a planoconvex lens.
(IV) First image has been viewed from the planar side of a planoconcave lens and second image from the convex side of a planoconvex lens. Which of the above statements are correct? (a) All statements are correct (b) Only statement III is correct (c) Only statement IV is correct (d) Only statements II, III and IV are correct
26. A ball is dropped vertically from height h and is bouncing elastically on the floor (see figure). Which of the following plots best depicts the acceleration of the ball as a function of time.
(B)
E
h
E (D)
(b) B
(c) C
t
(d) D
(a)
23. Ice in a freezer is at −7°C. 100 g of this ice is mixed
(b) 67 g
(c) 54 g
(d) 45 g
24. A point source of light is placed at 2f from a converging lens of focal length f. A flat mirror is placed on the other side of the lens at a distance d such that rays reflected from the mirror are parallel after passing through the lens again. If f = 30 cm, then d is equal to (a) 15 cm
(b) 30 cm
(c) 45 cm
(d) 75 cm
(b)
t
t
t
(d)
(c)
Acceleration
with 200 g of water at 15°C. Take the freezing temperature of water to be 0°C, the specific heat of ice equal to 2.2 J/g °C, specific heat of water equal to 4.2 J/g°C and the latent heat of ice equal to 335 J/g. Assuming no loss of heat to the environment, the mass of ice in the final mixture is closest to (a) 88 g
Second image
Acceleration
q
First image
Ignoring magnification effects, consider the following statements:
(III) First image has been viewed from the concave side of a planoconcave lens and second image from the planar side of a planoconvex lens.
q
q
KVPY
Acceleration
(b) 4 pπR 2
through different lenses such that board is at a distance beyond the focal length of the lens.
Acceleration
(a) 2 pπR 2
25. The word KVPY is written on a board and viewed
KVPY
20. In Guericke’s experiment to show the effect of
t
27. A student studying the similarities and differences between a camera and the human eye makes the following observations:
I. Both the eye and the camera have convex lenses.
35
KVPY Question Paper 2017 Stream : SA II. In order to focus, the eye lens expands or contracts while the camera lens moves forward or backward.
32. The stability of carbanions
III. The camera lens produces upside down real images while the eye lens produces only upright real images. IV. A screen in camera is equivalent to the retina in the eyes. V. A camera adjusts the amount of light entering in it by adjusting the aperture of the lens. In the eye, the cornea controls the amount of light. Which of the above statements are correct? (a) Statements I, II and IV are correct (b) Statements I, III and V are correct (c) Statements I, II, IV and V are correct (d) All statements are correct
σ
σ
CH3C(Ph)CH2CH3 IV
follows the order (a) III < IV < I < II (c) III < II < I < IV
(b) I < II < IV < III (d) IV < III < II < I
33. In the following reaction +
–
N2Cl
OH NaOH
+
the major product is OH O
velocity and comes to rest after moving distance d. During its motion, it had a constant acceleration f over 2/3 of the distance and covered the rest of the distance with constant retardation. The time taken to cover the distance is
(a)
(a) 2 d / 3 f
(c)
(c) 3d / f
CH3CHCH2CH3 II
(CH3)3C III
28. A particle starts moving along a line from zero initial
(b) 2 d / 3 f
σ
σ
CH3CH2CH2CH2 I
(d) 3 d / 2 f
(b) OH NH2
N N
(d)
29. If the image formed by a thin convex lens of power P has magnification m, then image distance v is (a) v =
1−m 1+ m m (b) v = (c) v = P P P
(d) v =
1 + 2m P
34. In the reaction of 1bromo3chlorocyclobutane with two equivalents of sodium in ether, the major product is
30. A long cylindrical pipe of radius 20 cm is closed at its upper end and has an airtight piston of negligible mass as shown. When a 50 kg mass is attached to the other end of the piston, it moves down. If the air in the enclosure is cooled from temperature T to T − ∆T , the piston moves back to its original position. Then ∆T / T is close to (Assuming air to be an ideal gas, g = 10 m/ s2, atmospheric pressure is 105 Pa)
(a) Br
(b) 0.02
(c) 0.04
(b) n
(c) Cl
Cl
(d)
35. The order of basicity of NH2
O2N
(a) 0.01
Cl
(d) 0.09
NH2
I
II
N III
N H IV
in water is
CHEMISTRY 31. The structure of 3methylpent2ene is
(a) IV < III < I < II (c) IV < I < III < II
(b) II < I < IV < III (d) II < III < I < IV
36. The first ionisation energy of Na, B, N and O atoms follows the order (a)
(b)
(c)
(d)
(a) B < Na < O < N (c) Na < O < B < N
(b) Na < B < O < N (d) O < Na < N < B
37. Among P2O5 , As2O3 , Sb2O3 and Bi2O3 , the most acidic oxide is (a) P2O5
(b) As 2O3
(c) Sb2O3
(d) Bi 2O3
KVPY Question Paper 2017 Stream : SA
36 38. Among K, Mg, Au and Cu, the one which is extracted by heating its ore in air is (a) K
(b) Mg
(c) Au
about the tobacco mosaic virus? (d) Cu
39. The metal ion with total number of electrons same as S2− is (a) Na +
(b) Ca 2 +
(c) Mg2 +
(d) Sr 2 +
40. X g of Ca [atomic mass = 40] dissolves completely in concentrated HCl solution to produce 5.04 L of H2 gas at STP. The value of X is closest to (a) 4.5
(b) 8.1
(c) 9.0
(d) 16.2
41. A 20 g object is moving with velocity 100 ms−1. The de Broglie wavelength (in m) of the object is [Planck’s constant h = 6.626 × 10−34 Js] (a) 3.313 × 10−34 (b) 6.626 × 10−34 (c) 3.313 × 10−31 (d) 6.626 × 10−31
42. In a closed vessel at STP, 50 L of CH4 is ignited with 750 L of air (containing 20% O2 ). The number of moles of O2 remaining in the vessel on cooling to room temperature is closest to (a) 5.8
(b) 2.2
(c) 4.5
(d) 6.7
43. CO2 is passed through lime water. Initially the solution turns milky and then becomes clear upon continued bubbling of CO2. The clear solution is due to the formation of (b) CaO (d) Ca(HCO3 )2
(a) CaCO3 (c) Ca(OH)2
44. The maximum number of electrons that can be filled in the shell with the principal quantum number n = 3 is (a) 18
(b) 9
(c) 8
(d) 2
45. The atomic radii of Li, F, Na and Si follow the order (a) Si > Li > Na > F (c) Na > Si > F > Li
(b) Li > F > Si > Na (d) Na > Li > Si > F
51. Which one of the following statements is correct about placenta? (a) Placenta is permeable to all bacteria (b) Oxygen and carbon dioxide cannot diffuse through the placenta (c) Waste products diffuse out of placenta into maternal blood (d) Placenta does not secrete chorionic gonadotropins
52. The respiratory quotient of the reaction given below is 2(C 51 H 98O6 ) + 145O2 → 102CO2 + 90H 2O + Energy (a) 0.703 (b) 0.725 (c) 0.960 (d) 1.422
53. Which one of the following statements is incorrect about nucleosomes? (a) They contain DNA (b) They contain histones (c) They are membranebound organelle (d) They are a part of chromosomes
54. The immediate precursor of thyroxine is (a) tyrosine (c) pyridoxine
(b) tryptophan (d) thymidine
55. The maximum number of oxygen molecules that can bind to one molecule of haemoglobin is (a) 8
(b) 6
(c) 4
(d) 2
56. Which one of the following biomolecules is synthesised in smooth endoplasmic reticulum? (a) Proteins (c) Carbohydrates
(b) Lipids (d) Nucleotides
include
46. The major excretory product of birds is (b) uric acid (d) ammonia
47. Codon degeneracy means that (a) several amino acids are coded by more than one codon (b) one codon can code for many amino acids (c) one amino acid can be coded by only one codon (d) the codons are triplet nucleotide sequences
48. In cell cycle, during interphase (a) two daughter cells are produced (b) the nucleus is divided into two daughter nuclei (c) the chromosome condenses (d) the DNA is replicated
49. Transfer of genetic material between population is best defined as (a) gene flow (c) genetic shift
(a) It affects all monocotyledonous plants (b) It affects photosynthetic tissue of the infected plant (c) It does not infect other species belonging to the Solanaceae (d) It infects gymnosperms
57. The products of light reaction during photosynthesis
BIOLOGY (a) urea (c) nitrates
50. Which one of the following statements is correct
(b) genetic drift (d) speciation
(a) ATP and NADPH (c) O2 and H2O
(b) O2 and NADP + (d) NADP + and H2O
58. Hypothalamus directly controls the production of which of the following hormones? (a) Glucocorticoid and insulin (b) Insulin and glucagon (c) Atrial natriuretic factor and gastrin (d) Glucocorticoids and androgens
59. Which one of the following drug is not obtained from fungal or plant sources? (a) Penicillin (c) Acetaminophen
(b) Reserpine (d) Quinine
60. Jean Baptiste Lamarck explained evolution based on (a) natural selection (b) survival of the fittest (c) mutations (d) inheritance of acquired characteristics
37
KVPY Question Paper 2017 Stream : SA
PARTII (2 Marks Questions) MATHEMATICS
Choose the correct option.
61. Let S be the circle in XY plane which touches the Xaxis at point A, the Y axis at point B and the unit circle x2 + y2 = 1 at point C externally. If O denotes the origin, then the angle OCA equals π (b) 2
3π (c) 4
(b) CL1 > CL 2 and U1 > U 2 (d) CL1 < CL 2 and U1 < U 2
67. A long horizontal mirror is next to a vertical screen (seen figure).
3π (d) 4
Screen
5π (a) 8
(a) CL1 > CL 2 and U1 < U 2 (c) CL1 < CL 2 and U1 > U 2
62. In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the nonparallel sides are all equal to 30.In order to maximise the area of the trapezium, the smallest angle should be (a)
π 6
(b)
π 4
(c)
π 3
(d)
π 2
63. Let A1 , A2 , A3 be regions in the XY plane defined by A1 = {(x, y) : x2 + 2 y2 ≤ 1} A2 = {(x, y) :x3 + 2 2  y3 ≤ 1} A3 = {(x, y) : max ( x, 2  y) ≤ 1} Then, (a) A1 ⊃ A2 ⊃ A3 (c) A2 ⊃ A3 ⊃ A1
(b) A3 ⊃ A1 ⊃ A2 (d) A3 ⊃ A2 ⊃ A1
ABCD such that E , A, C are collinear in that order. Suppose EB = ED = 130 and the areas to ∆EAB and square ABCD are equal. Then, the area of square ABCD is (b) 10
(c) 120
d
Parallel light rays are falling on the mirror at an angle α from the vertical. If a vertical object of height h is kept on the mirror at a distance (d > h ) tan α. The length of the shadow of the object on the screen would be h 2 (c) 2h
(a)
64. Let ABCD be a square and E be a point outside
(a) 8
α
h
(b) h tanα (d) 4h
68. A spherical marble of radius 1 cm is stuck in a circular hole of radius slightly smaller than its own radius (for calculation purpose, both can be taken same) at the bottom of a bucket of height 40 cm and filled with water up to 10 cm.
(d) 125
65. Consider the set A = {1, 2, 3, ... , 30}. The number of ways in which one can choose three distinct number from A so that the product of the chosen numbers is divisible by 9 is (a) 1590
(b) 1505
(c) 1110
(d) 1025
PHYSICS
If the mass of the marble is 20 g, then the net force on the marble due to water is close to (a) 0.02 N upwards (c) 0.04 N upwards
(b) 0.02 N downwards (d) 0.04 N downwards
69. In the circuit shown below (on the left) the resistance
66. Two different liquids of same mass are kept in two identical vessels, which are placed in a freezer that extracts heat from them at the same rate causing each liquid to transform into a solid. The schematic figure below shows the temperature T versus time t plot for the two materials. We denote the specific heat in the liquid states to be C L1 and C L 2 for materials 1 and 2, respectively and latent heats of fusion U 1 and U 2, respectively. T
and the emf source are both variable. R
A V
V 2V0
H F
V0
1
G E
2
0
t
0
I0
2I0
I
KVPY Question Paper 2017 Stream : SA
38 The graph of seven readings of the voltmeter and the ammeter (V and I, respectively) for different settings of resistance and the emf, taken at equal intervals of time ∆t, are shown below (on the right) by the dots connected by the curve EFGH. Consider the internal resistance of the battery to be negligible and the voltmeter an ammeter to be ideal devices. (Take, V R0 ≡ 0 ). I0 Then, the plot of the resistance as a function of time corresponding to the curve EFGH is given by
CHO,
CHO
Y= Ph
O
O
(b) X = H3C
CH3 ,
Y=
CH3 ,
Y=
H3C
Ph
O H 3C (c) X = H3C
Ph CH3
H F
R0
G E
0
2∆t
4∆t
R0
2∆t
4∆t
6∆t t
2∆t
4∆t
(a) 2.5
E
R0 /2 0 6∆t t 0
F
(b) 5
(c) 1.25
(d) 2
H
metals is shown in the plot below. If light of wavelength 400 nm is incident on each of these metals, which of them will emit photoelectrons?
6∆t t
[Planck’s constant h = 6626 × 10−34 Js; velocity of . 8 −1 light c = 3 × 10 m s ; 1 eV = 16 . × 10−19 J]
G 2∆t
4∆t
70. Stokes’ law states that the viscous drag force F experienced by a sphere of radius a, moving with a speed v through a fluid with coefficient of viscosity η, is given by F = 6πηav. If this fluid is flowing through a cylindrical pipe of radius r, length l and pressure difference of p across its two ends, then the volume of water V which flows through the pipe in time t can be a V p written as = k ηbr c, where k is a dimensionless l t constant. Correct values of a , b and c are (a) a = 1, b = − 1, c = 4 (c) a = 2, b = − 1, c = 3
number of moles of oxygen produced per mole of KMnO4 is
74. The photoelectric behaviour of K, Li, Mg and Ag R0
G
Ph
73. KMnO4 reacts with H2O2 in an acidic medium. The
H F
Y = H3C
H
(d) R
E
CHO ,
(d) X = H3C
F G
R0 /2 0 0 6∆t t
R 2R0
R0 /2 0 0
E
(b) a = − 1, b = 1, c = 4 (d) a = 1, b = − 2, c = − 4
(a) K (c) K, Li and Mg
(b) K and Li (d) K, Li, Mg and Ag
75. A piece of metal weighing 100 g is heated to 80°C and dropped into 1 kg of cold water in an insulated container at 15°C. If the final temperature of the water in the container is 15.69°C, the specific heat of the metal in J/g.°C is (a) 0.38 (c) 0.45
(b) 0.24 (d) 0.13
BIOLOGY 76. The nucleus of a diploid organism contains 3 ng of
CHEMISTRY 71. The reaction of an alkene X with bromine produces a compound Y, which has 22.22% C, 3.71% H and 74.07% Br. The ozonolysis of alkene X gives only one product. The alkene X is, [Given, atomic mass of C = 12; H = 1; Br = 80] (a) ethylene (c) 2butene
(b) 1butene (d) 3hexene
72. In the following reaction, Hg 2 +
DNA in G1 phase. Which one of the following statements describes the state of the cell at the end of Sphase? K
Li
Mg
kinetic energy of photoelectron
R0
(c)
(a) X = H3C
(b) R
(a) R 2R0
0
H 3C
dil. NaOH
H3 C — C ≡≡ C — H → X → Y +
H 3O
X and Y, respectively, are
PhCHO
2 3 4 5 Incident photon energy (eV)
Ag
39
KVPY Question Paper 2017 Stream : SA
78. The concentration of OH− ions in a solution with the
(a) The nucleus divides into two and each nucleus contains 3 ng of DNA (b) The nucleus does not divide and it contains 3 ng of DNA (c) The nucleus divides into two and each nucleus contains 1.5 ng of DNA (d) The nucleus does not divide and it contains 6 ng of DNA
H+ ions concentration of 13 . × 10−4 M is
(a) 7.7 × 10−4 M −8
(c) 2.6 × 10
(d) 7.7 × 10−11 M
M
79. Given that tidal volume is 600 mL, inspiratory reserve volume is 2500 mL and expiratory reserve volume is 800 mL, what is the value of vital capacity of lung?
77. Three cellular processes are listed below. Choose the correct combination of processes that involve proton gradient across the membrane.
(a) 3900 mL (c) 3100 mL
(b) 3300 mL (d) 1400 mL
80. Which of the following organisms produces sperm
I. Photosynthesis II. Aerobic respiration III. Anaerobic respiration (a) II and III (c) I, II and III
(b) 1.3 × 10−4 M
without involving meiosis? (a) Sandfly and fruitfly (c) Honeybee and ant
(b) I and II (d) I and III
(b) Housefly and grasshopper (d) Zebra fish and frog
Answers PARTI 1
(b)
2
(c)
3
(b)
4
(b)
5
(d)
6
(c)
7
(d)
8
(d)
9
(b)
10
(c)
11
(a)
12
(a)
13
(b)
14
(d)
15
(d)
16
(c)
17
(d)
18
(c)
19
(b)
20
(c)
21
(a)
22
(c)
23
(b)
24
(c)
25
(d)
26
(b)
27
(c)
28
(c)
29
(a)
30
(c)
31
(a)
32
(c)
33
(d)
34
(d)
35
(c)
36
(b)
37
(a)
38
(d)
39
(b)
40
(c)
41
(a)
42
(b)
43
(d)
44
(a)
45
(d)
46
(b)
47
(a)
48
(d)
49
(a)
50
(b)
51
(c)
52
(a)
53
(c)
54
(a)
55
(c)
56
(b)
57
(a)
58
(d)
59
(c)
60
(d)
PARTII 61
(a)
62
(c)
63
(d)
64
(b)
65
(a)
66
(c)
67
(c)
68
(*)
69
(d)
70
(a)
71
(c)
72
(b)
73
(a)
74
(b)
75
(c)
76
(d)
77
(b)
78
(d)
79
(a)
80
(c)
KVPY Question Paper 2017 Stream : SA
40
Solutions 8π 2π 4π × 4= ⇒ ∠A2OA4 = 9 9 9 OA1 = OA2 = r In ∆A1 OA2,
∴ ∠A1OA5 =
1. (b) We have, side of quadrilateral has distinct integer second largest size has length 10. Let a = 8, b = 9, c = 10, (All are distinct) We know, in quadrilateral Sum of three sides is greater than fourth side ∴ a + b + c > d ⇒ 8 + 9 + 10 > d ⇒ d < 27 ∴ Maximum length of 4th side is 26.
A6 A5 A7
2. (c) Exponent of 2 in 200!. A4
200 200 200 200 200 + + + = + 2 22 23 24 25 200 200 200 + 6 + 7 + 8 2 2 2 = 100 + 50 + 25 + 12 + 6 + Exponent of 2 in 100! 100 100 100 + + + = 2 22 23
2π 9
3 + 1 = 197
100 + 100 24 25 100 100 + 6 + 7 2 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97 ∴Exponent of 2. 200! 2197 In = = 2100 100! 297
A1
cos
2
1 1 1 1 1 1 + + + − + + − − 2 2 2 2 2 2 0 + 1+ 0 + 1= 2 The value lies between (1.5, 2.5). 1− 2
1 2
4. (b) x and y are positive integer x2 − y2 = 12345678 RHS 12345678 is and even number and last digit is 8. ∴ The last digit of x be 3, 7 and the last digit of y be 1, 9. ∴x and y must be odd and square of difference is multiple of 8 but RHS is not multiple of 8. ∴ S is the empty set.
5. (d) Given, A1 , A2 , A3 , ..., A9 are nineside regular polygon of each side 2 units. ∴ A1 A2 = A2A3 = A3 A4 = ..... = A8 A9 = 2 2π ∠A1OA2 = 9
⇒ n 2 − 100 − n − 110 = 0 [Q P (n − P ) = 50] A3
A2
2 π OA12 + OA22 − A1 A22 = 9 2OA1OA2
2 π 2r 2 − 4 = 9 2r 2 2π = r2 − 2 r 2 cos 9 2 2 r2 = = 2π π 1 − cos 2 sin 2 9 9 1 r= π sin 9 8 π r 2 + r 2 − A1 A52 In ∆A1OA5 , cos = 9 2r 2 4π A1 A5 = 2r sin ⇒ 9 Similarly, ∆A2OA4 , 2π A2A4 = 2r sin 9 4π 2π ∴ A1 A5 − A2A4 = 2r sin − sin 9 9 π π = 2r, 2 sin cos 9 3 1 π 1 2 = × 2 sin × Q r = π π 9 2 sin sin 9 9 cos
and a12 + a22 + a32 + a42 = 1 It is possible only 1 1 when, a1 = a2 = and a3 = a4 = − 2 2 ∴ (a1 − a2 )2 + (a2 − a3 )2 + (a3 − a4 )2 + (a4 − a1 )2 2
4π 9
A9
∴ The largest power of 2 is 100. 3. (b) Given, a1 + a2 + a3 + a4 = 0
2
O
A8
2
∴ a j and ak are both positive or negative. P ∴ C2 + n − P C2 = 55 and a j ak is negative j < k a j ak = 50 any one of a j and ak are positive: P ∴ C1 × n − P C1 = 50 ⇒ P (n − P ) = 50 P ⇒ C2 + n − P C2 = 55 ⇒ P (P − 1) + (n − P ) (n − P − 1) = 110 ⇒ P 2 − P + (n − P )2 − (n − P ) = 110 ⇒ P 2 + (n − P )2 − n = 110 ⇒ {P + (n − P )}2 − 2P (n − P ) − n = 110
=2
∴ n 2 − n − 210 = 0 ⇒ (n − 15) (n + 14) = 0 n = 15, n ≠ − 14 ∴ P (15 − p ) = 50 ⇒ p 2 − 15 p + 50 = 0 (P − 10) (P − 5) = 0, p = 5 or 10 ∴ p 2 + (n − P )2 = 52 + 102 = 25 + 100 = 125
7. (d) We have, a , b, c, d are four distinct number from the set {1, 2, 3, ..., 9}. a c The minimum value of + is possible b d when a = 2, b = 9, c = 1, d = 8 2 1 16 + 9 25 + = = ∴ 9 8 72 72 8. (d) Given, 72x ⋅ 48y = 6xy (23 ⋅ 32 )x ⋅ (24 ⋅ 3)y = 2xy ⋅ 3xy 23 x + 4 y ⋅ 32x + y = 2xy ⋅ 3xy Equating the exponent of 2 and 3, we get 3x + 4 y = xy and 2x + y = xy On solving these equation, we get −15 5 and y = x= 3 3 − 15 5 − 10 ∴ x+ y= + = 3 3 3
9. (b) Given, AB is diameter of circle S and C is the midpoint of arc length of AB. T
C 3
6. (c) Let p are positive number from a1 , a2 , a3 , …, an ∴ n − p are negative number. Given a j , ak is positive j < k and a j ak = 55 a j ak is positive.
√2 A
1
1
O
1
B
41
KVPY Question Paper 2017 Stream : SA AC is diameter of circle T . AB = 2 ∴ OA = OB = OC = 1 Area of shaded region = Area of semicircle T + Area of ∆OAC − Area of quadrant of circle S π 2 π = 4 =
2
2 1 π 2 + × 1 × 1 − × (1) 2 2 4 1 π 1 + − = 2 4 2
10. (c) Let p(x) = x135 + x125 − x115 + x5 + 1, q(x) = x3 − x and p (x) = q(x)k + r (x) x135 + x125 − x115 + x5 + 1 = (x3 − x)k + ax2 + bx + c [Q r (x) = ax2 + bx + c] Put x = 0, ∴ c=1 Put x = 1, 3 = a + b + c ⇒ 3 = a + b + 1 … (i) ⇒ a+ b= 2 Put x = − 1, − 1 = a − b + c ⇒ − 1 = a − b + 1 … (ii) ⇒ a− b= − 2 From Eqs. (i) and (ii), we get a = 0, b = 2 ∴ r (x) = 2x + 1 ∴ Degree of r (x) = 1
11. (a) The distinct prime factor of 43361 = 131 × 331 where p1 = 131 and p2 = 331 ∴ p1 + p2 = 131 + 331 = 462
12. (a) Given, ABC is right angled triangle ∠C = 90° CD is perpendicular on AB, DN and DM are parallel to AC and BC, respectively. DN = 4 and DM = 5
In ∆DNC and ∆DMA, ∆DNC ~ ∆DMA DN NC = ⇒ DM MA 25 4 5 ⇒ = ⇒ MA = 4 5 MA 25 41 ∴ AC = MC + AM = 4 + = 4 4 13. (b) We have, A , G , H be arithmetic, geometric and harmonic mean respectively of two distinct positive real numbers. A (G − H )x2 + G (H − A )x + H (A − G ) = 0 Let α and β be roots of the given equation α G ]
α F. 46. (b) The major excretory product of birds is uric acid. Nitrogenous wastes in the body of animals tend to form toxic ammonia, which must be excreted. Mammals such as humans excrete urea, while birds, reptiles and some terrestrial invertebrates produce uric acid as waste excretory product.
47. (a) There are 64 codons present in each living organism, out of which, 61 codons represent or code for amino acids and rest three are stop codons. Thus, there are more codon combinations than there are amino acids. The genetic code is described as degenerate because more than one codon sequence can code for the same amino acid. 48. (d) In cell cycle, during interphase, the DNA is replicated. Interphase begins with G1 phase. During this phase, the cell makes a variety of proteins that are needed for DNA replication. G1 phase is followed by synthetic or Sphase. This phase is responsible for the synthesis or replication of DNA. The aim of this process is to produce double the amount of DNA, providing the basis for the chromosome sets of the daughter cells.
49. (a) Gene flow is the transfer of genetic variation from one population to another. Genetic drift is a change in the frequency of an allele within a population over time. Speciation is the formation of new and distinct species in the course of evolution. Genetic shift is a major change within a population which changes the population altogether.
50. (b) Tobacco Mosaic Virus (TMV) affects photosynthetic tissue of the infected plant. Other statements can be corrected as TMV affects all dicotyledonous plants, of which most important are tobacco and tomato. But it does not affect any monocotyledonous plant. TMV is a ssRNA virus, it infects a wide range of plants, especially tobacco and other members of the family Solanaceae. TMV does not infect gymnosperms.
51. (c) Placenta allows the foetus to transfer waste products to the mother’s blood. Other statements can be corrected as Placenta gives protection against most bacteria and does not allow infections to enter the foetus. Placenta allows gaseous exchange via the mother’s blood supply, i.e. it allows diffusion of O2 and CO2 . Placenta produces hormones like human Chorionic Gonadotropin (hCG), progesterone, oestrogen and human Placental Lactogen (hPL).
52. (a) Respiratory Quotient (RQ) measures the ratio of the volume of carbon dioxide (VC ) produced by an organism to the volume of oxygen consumed (VO ). The RQ for the given equation is CO2 produced 102 RQ = = = 0.703 O2 consumed 145 RQ = 0.703
53. (c) Nucleosome is a structural unit of a eukaryotic chromosome, consisting of a length of DNA coiled around a core histone. Thus, nucleosome is not a membrane bound organelle of a cell.
54. (a) Tyrosine is the immediate precursor of the thyroxine hormone. Thyroxine is produced in the thyroid gland from tyrosine and iodine. Thyrotropin Releasing Hormone (TRH) is produced by the hypothalamus which induces the release of thyroxine. 55. (c) The haemoglobin molecule has four binding sites for oxygen molecules. Thus, each Hb tetramer can bind four oxygen molecules. Haemoglobin is the oxygen transporting protein of red blood cells and is a globular protein with quaternary structure. Haemoglobin consists of four polypeptide subunits, 2α chains and 2β chains.
45
KVPY Question Paper 2017 Stream : SA 56. (b) The smooth endoplasmic reticulum functions in lipid synthesis and metabolism, the production of steroid hormones and detoxification. Smooth Endoplasmic Reticulum (SER) is a meshwork of five disclike tubular membrane vesicles, part of a continuous membrane organelle within the cytoplasm of eukaryotic cells.
during his life in order to adapt to its environment, those changes are passed on to its offsprings.
61. (a) OC = 1radius of circle x2 + y2 = 1 OA = AP ∠AOP = ∠OPA = 45°
∴
photosynthesis uses light energy to make two molecules needed for the next stage of photosynthesis. These include the energy storage molecule ATP and the reduced electron carrier NADPH. The light reaction takes place in the thylakoid membranes of chloroplasts.
58. (d) Hypothalamus directly controls the production of glucocorticoids and androgens. These hormones are secreted in response to ACTH (Adrenocorticotropic Hormone), which is secreted from the anterior pituitary gland. The ACTH is released in response to corticotropin releasing hormone from the hypothalamus. The pathway can be explained as Hypothalamus ACTHreleasing hormone
63. (d) Given, A1 = {(x, y) : x2 + 2 y2 ≤ 1} A2 = {(x, y) :x3 + 2 2y3 ≤ 1} A3 = {(x, y) : max (x, 2 y1 ≤ 1)} Graph of A1 , A2 and A3 are
P
B
57. (a) The light dependent reaction of
C O
A
1 y = √2
x=–1
( 0, √12 ( x2+2y2=1
AP = PC radius of circle In ∆PCA, ∴∠PCA + ∠PAC + ∠CPA = 180° ⇒ 2∠PCA + 45° = 180° 135° ∠PCA = ⇒ 2 ⇒ ∠OCA = 180° − ∠PCA 135° 3π ∠OCA = 180° − = π− ⇒ 2 8 5π ⇒ ∠OCA = 8
–1, 0
D
(1, 0)
( 0, – √12 (
Clearly from graph A1 ⊂ A2 ⊂ A3
64. (b) Given, area of ∆EAB = area of square ABCD EB = ED = 130 Let side of square = x x BM = = AM 2
62. (c) Given ABCD is trapezium
ACTH
D
C
C 30
30 Adrenal gland
M
θ A Glucocorticoids
Androgens
59. (c) Acetaminophen, also known as paracetamol is not produced by plant or fungi, it is artificially formed. The starting material for the manufacturing of paracetamol is phenol which is nitrated to give a mixture of the ortho and paranitrotoluene. Other drugs are obtained as, penicillin is an antibiotic obtained from ascomycetous fungi Penicillium notatum. Reserpine is an alkaloid derived from the roots of Rauwolfia serpentina plant. Quinine comes from the bark of the Cinchona tree.
60. (d) Lamarck is best known for his theory of Inheritance of Acquired Characteristics’, first presented in 1801. It states that if an organism changes
x=1
Max(x, 2√2/y)≤1)
y=– 1 √2
AD = BC = CD = 30 Let the smallest angle be θ. Pituitary gland
3θ π θ π = or = 2 2 2 2 π θ = or θ = π ⇒ 3 π For maximum θ = 3 ⇒
M
N
x √2
x
B
∠DAB = θ
A
x
B
In ∆AMD, AM ⇒ AM = 30cosθ 30 DM sin θ = = DM = 30 sin θ 30 1 Area of trapezium = (AB + CD ) DM 2 1 A = (60 + 60 cosθ) 30 sin θ ⇒ 2 ⇒ A = 900 (sin θ + sin θ cosθ) dA = 900 (cosθ − sin 2 θ + cos2 θ) ⇒ dθ dA For maximum or minimum, put =0 dθ ∴ cosθ − sin 2 θ + cos2 θ = 0 ⇒ cosθ + cos 2θ = 0 3θ θ 2 cos cos = 0 ⇒ 2 2 cosθ =
E
Area of ∆AEB = Area of ∆BEM − area of (∆AMB ) 1 1 = EM × BM − AM × BM 2 2 1 = BM (EM − AM ) 2 x2 x 1 x 130 − = − 2 2 2 2 ∴ ⇒
x2 x 1 x = x2 . − 130 − 2 2 2 2 130 −
x2 x = 2 2x + 2 2
KVPY Question Paper 2017 Stream : SA
46 130 −
⇒
x2 5x = 2 2
From similar triangles ∆BGF and ∆DEF,
2
DE FE = BG GF
we have
x2 25 x2 ⇒ = 130 − 2 2 ⇒ 13x2 = 130 ⇒ x2 = 10 ∴Area of square = 10
d h′ + h …(i) = x h Now, from similar triangles ∆ABG and ⇒
Case I All three number are multiple of 3 then product of three number are divisible by 9. 10 ∴ C3 = 120 Case II Two number are multiple of 3 and other are not multiple of 9. 10 i.e. C2 × 20C1 = 900 Case III One are multiple of 9 and other two are not multiple of 3. 3 C1 × 20C2 = 570 ∴Total number of ways = 120 + 900 + 570
∆ACE, we have CE BG = AE AG ∴
AG = GF = x
and ⇒
H + h′ h = d+ x x
66. (c) We have, heat extracted from a liquid during solidification, U = Qt = mL ⇒ L ∝ U Also, heat extracted from liquid during cooling, H = Qt = mc∆T Temperature of liquid, Q T = ⋅ t + Ti mc Slope of T versus t line is inversely proportional to specific heat c. Now, from given graph, we get T 1
Slope of 2 is more ⇒ CL1 U 2 and CL1 < CL 2
67. (c) From geometry of figure, shadow length is CD (= H ). C
In be cid a e lig m o nt ht f
h′ + h H + h′ − h = h h ⇒ 2h = H Hence, height of shadow on wall is 2h.
68. (No option is matching) Net force on marble due to water is
hα
α α
90–α 90–α
A
G x
F
Fnet = (Force of water column of height ~ 9 cm) − (Buoyant force on marble) = πr 2 ρw gh − Volume of marble under water × ρw × g 2 = πr 2ρw gh − πr3 ρw g 3 2 2 = πr ρw g h − r 3 ≈ 314 . × (1 × 10−2 )2 × 1000 × 10 9 − 2 × 10−2 3 = 0.26 N (downwards)
V 2V0
a
ML−1 T −2 c −1 −1 b [L3 T −1 ] = [ML T ] [L] L Equating powers of M, L and T, we get a + b = 0 ⇒ − 2a − b + c = 3 −2a − b = − 1 Solving, we get a = 1, b = − 1 and c = 4
X + Br2
V0
F
I0
2I0
One product
Emperical formula for Y can be calculated as, Elements
C H Br
Moles Simplest Simplest % of At whole molar of elem mass no. ratio element ent
22.22/12 1.85/0.92 2 × 2
22.22% 12 3.71%
1
= 1.85
= 2.01
3.71/1
3.71/0.92 4 × 2
= 3.71
= 4.03
74.07/80 0.92/0.92 1 × 2
74.07% 80
= 0.92
=1
∴The emperical formula of Y is C4 H8 Br2. According to retero synthesis. CH
CH — CH3+Br2
(X) 2butene
CH3
G
E
Y
O3/Zn, H2O
CH3
H
E d –x
d
h′
90–α
p V = k ηb r c , l t
we have
1 cm
= 26 × 10
D
a
70. (a) From
9 cm
69. (d) In given V I graph,
B
V = V 0 = constant But current increases, so resistance must decreases. R V At G, resistance = 0 = 0 . 2 2I 0
71. (c) Skeletal diagram for the given information can be shown as:
H
90–α
…(ii)
Equating Eqs. (i) and (ii), we get
−2
α
From F to G,
So, correct option is (d).
d H + h′ − h = x h
⇒
V0 I0
From G to H, current is constant but voltage increases, so resistance decreases. 2V 0 V 0 At H, resistance = = = R0 2I 0 I0
∆ABG ≅ ∆FBG
= 1590
U1>U2
∴ At F, resistance = R0 =
h′ d − x = h x
65. (a) Given, A = {1, 2, 3, …, 30}
From E to F, V 0 = I 0 R0 s lope = R0 ⇒
I
CH
CH
Br
Br
CH3
(Y ) 2, 3dibromobutane
47
KVPY Question Paper 2017 Stream : SA Also, CH3
CH
CH (X)
CH3 + O3 O CH3
CH
CH
O
CH3
= 400 × 10 hc Energy of photon E = λ
Zn/H2O
E=
2CH3C H Acetaldehyde
C
H
Hg2+ H3O+
H3C
C
PhCHO
H3C (X) dil. NaOH (aldol condensation)
O CH3
C
CH
CH
Ph
(Y) (α, βunsaturated ketone)
In first step one molecule of water adds to alkyne on warming with mercuric sulphate and dilute sulphuric acid to form carbonyl compound, i.e. acetone. In second step 2 molecules of acetone condense in presence of dil. NaOH to form α , βunsaturated ketone. This reaction is known as aldol condensation.
73. (a) For the reaction, +7
6.626 × 10−34 Js × 3 × 108 m / s 400 × 10−9 m
= 4.97 × 10−19 × O
C
m
E = 4.97 × 10−19 J
72. (b)
CH3
74. (b) Given, wavelength λ = 400 nm −9
O
O
∴ (Number of eq.)KMnO4 = (Number of eq.)H 2O2 1× 5 = x × 2 5 x = = 2. 5 2
−4
2MnO−4 + 5H2 O 2 + 6H+ → 2Mn
2+
−2
+ 8H2 O + 5O2
Let the number of moles of oxygen produced per mole KMnO4 be x. Number of equivalent = Number of moles × change in oxidation state Number of equivalent of KMnO4 = 1 × 5 Number of equivalent of H 2O2 = x × 2
1 eV 1.6 × 10−19 J
E = 31 . eV If the energy of incident light ≥ work function of light. Then photoelectrons will be ejected. Thus, K and Li will emit photoelectrons as their threshold energy (obtained from graph) is less than 3.1 eV.
75. (c) Given, weight of metal = 100 g T2 = 15.69° C, T1 = 80° C Specific heat of water = 4184 J/g°C . Heat gained by 100 g of metal = Heat lost by 1000 g of water. We know, Q = mc ∆T ∴ 100 × x × (80 − 15.69) = 1000 × 4184 . (15. 69 − 15) = 100 x (64.31) = 2886.96 2886.96 x= = 0.448 ≈ 0.45 J/g.°C 6431
76. (d) In the G1 phase of cell cycle, the cell grows in size, i.e. the cell synthesises various enzymes and nutrients that are needed later on for DNA replication. The next phase of cell cycle is Sphase during which the DNA amount doubles up, i.e. a
cell with 3 ng of DNA in G1 phase will now have 6 ng of DNA. G 2 phase comes after Sphase. It is second growth phase but here the DNA content will remain 6 ng.
77. (b) In photosynthesis, photophosphorylation and in aerobic respiration, oxidative phosphorylation occurs that requires proton gradient. The sunlightdriven production of ATP from ADP and inorganic phosphate is called photophosphorylation. It occurs in the chloroplast. Oxidative phosphorylation is the process in which ATP is formed by the transfer of electrons from NADH or FADH 2 to O2 by a series of electron carriers. It occurs inside the mitochondria. 78. (d) [H + ] [OH − ] = 10−14 1.3 × 10−4 × [OH − ] = 10−14 1 [OH − ] = × 10+4 × 10−14 1.3 1 = × 10−10 = 0.769 × 10−10 1.3 = 0.77 × 10−10 = 7.7 × 10−11 M
79. (a) Vital capacity = Inspiratory reserve volume + Tidal volume + Expiratory reserve volume = 2500 mL + 800 mL + 600 mL = 3900 mL Vital capacity is the volume of air breathed out after the deepest inhalation. 80. (c) All haploid sexually reproducing organisms would produce sperms/male gametes without the process of meiosis, e.g. Honeybee (Apis) and Ant (Formica). Haploid parents produce gametes by mitotic division. This happens because meiosis is reductional division in which the daughter cells contain half the number of chromosomes as the parent cell. Therefore, haploid organisms do not show meiosis to further disturb their ploidy.
KVPY Question Paper 2016 Stream : SA
48
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2016 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS
5. Let a1 , a 2 , ... , a100 be nonzero real numbers such that 2
1. Suppose the quadratic polynomial P (x) = ax + bx + c has positive coefficients a , b, c in arithmetic progression in that order. If P (x) = 0 has integer roots α and β. Then, α + β + αβ is equal to (a) 3
(b) 5
(c) 7
(d) 14
2. The number of digits in the decimal expansion of 5 16
16 5
is
(a) 16
(b) 17
(c) 18
(d) 19
3. Let t be real number such that t 2 = at + b for some positive integers a and b. Then, for any choice of positive integers a and b, t3 is never equal to (a) 4t + 3
(b) 8t + 5
(c) 10t + 3
(d) 6t + 5
4. Consider the equation (1 + a + b)2 = 3(1 + a 2 + b2) , where a , b are real numbers. Then, (a) there is no solution pair (a , b) (b) there are infinitely many solution pairs (a , b) (c) there are exactly two solution pairs (a , b) (d) there is exactly one solution pair (a , b)
a1 + a 2 + ... + a100 = 0 Then, ai − ai 100 (a) Σ100 0 and Σi = 1 ai 2 ai − ai 100 (b) Σ100 ≥0 i = 1 ai 2 ≥ 0 and Σi = 1 ai 2 ai 100 − ai (c) Σ100 ≤0 i = 1 ai 2 ≤ 0 and Σi = 1 ai 2 ai − ai 100 depends on the (d) The sign of Σ100 i = 1 ai 2 or Σi = 1 ai 2 choice of ai ’ s
6. Let ABCD be a trapezium, in which AB is parallel to CD, AB = 11, BC =4, CD = 6 and DA = 3. The distance between AB and CD is (a) 2 (b) 2.4 (c) 2.8 (d) Not determinable with the data
7. The points A, B, C , D, E are marked on the circumference of a circle in clockwise direction such that ∠ABC = 130° and ∠CDE = 110°. The measure of ∠ACE in degree is (a) 50° (c) 70°
(b) 60° (d) 80°
49
KVPY Question Paper 2017 Stream : SA 8. Three circles of radii 1, 2 and 3 units respectively touch each other externally in the plane. The circumradius of the triangle formed by joining the centers of the circles is (a) 1.5
(b) 2
(c) 2.5
(d) 3
9. Let P be a point inside a ∆ ABC with ∠ABC = 90°. Let P1 and P2 be the images of P under reflection in AB and BC respectively. The distance between the circumcenters of ∆ ABC and P1PP2 is AP + BP + CP 3 AB + BC + AC (d) 2
AB 2 AC (c) 2
(b)
(a)
10. Let a and b be two positive real numbers such that
a + 2b ≤ 1. Let A1 and A2 be respectively the areas of circles with radii ab3 and b2. Then, the maximum A possible value of 1 is A2
1 16 1 (c) 16 2
1 64 1 (d) 32
(b)
(a)
(c) 135
(d) 150
12. Consider a cuboid all of whose edges are integers and whose base is a square. Suppose the sum of all its edges is numerically equal to the sum of the areas of all its six faces. Then, the sum of all its edges is (a) 12
(b) 18
(c) 24
far would she have to walk due north and then due east to arrive at the same location? (a) Towards north 2.88 km and towards east 1.34 km (b) Towards north 2.11 km and towards east 2.11 km (c) Towards north 1.25 km and towards east 1.93 km (d) Towards north 1.34 km and towards east 2.88 km
17. The length and width of a rectangular room are measured to be 395 . ± 005 . m and 305 . ± 005 . m, respectively. The area of the floor is
(a) 12.05 ± 0.01 m2 (c) 12.05 ± 0.34 m2
(d) 36
13. Let A1 , A2 ,..... , Am be nonempty subsets of { 1, 2, 3, …, 100} satisfying the following conditions: 1. The numbers A1, A2, K , Amare distinct. 2. A1 , A2 ,... , Am are pairwise disjoint. (Here Adonotes the number of elements in the set A)
18. A car goes around uniform circular track of radius R at a uniform speed v once in every T seconds. The magnitude of the centripetal acceleration is a c. If the car now goes uniformly around a larger circular track of radius 2R and experiences a centripetal acceleration of magnitude 8a c. Then, its time period is (b) 3T
(b) 14
(c) 15
(c) T /2
(d) 3/2T
contain 10 and 100 turns, respectively. The primary coil is connected to a battery that supplies a constant voltage of 1.5 V. The voltage across the secondary coil is (a) 1.5 V
(b) 0.15 V
(c) 0.0 V
which generates electricity. How much water must fall per second in order to generate 100 . × 109 W of power ? (Assume 50% efficiency of conversion and g =10 ms−2) (a) 250 m3 (c) 500 m3
(b) 400 m3 (d) 200 m3
21. The diagram below shows two circular loops of wire (A and B) centred on and perpendicular to the Xaxis and oriented with their planes parallel to each other. The Y axis passes vertically through loop A (dashed line). There is a current I B in loop B as shown in the diagram. Possible actions which we might perform on loop A are Y
(d) 16 IB
14. The number of all 2digit numbers n, such that n is equal to the sum of the square of digit in its tens place and the cube of the digit in units place is (a) 0
(b) 1
(c) 2
integers such that f (xy) = f (x) + f ( y) for all positive integers x, y. If f (12) = 24 and f (8) = 15. The value of f (48) is (b) 32
(c) 33
–X
X
(d) 4
15. Let f be a function defined on the set of all positive
(a) 31
(d) 15 V
20. Water falls down a 500.0 m shaft to reach a turbine
Then, the maximum possible value of m is (a) 13
(b) 12.05 ± 0.005 m2 (d) 12.05 ± 0.40 m2
19. The primary and the secondary coils of a transformer
Both of them burn at uniform rate. The first one burns in 5 hr and the second one burns in 3 h. Both the candles are lit together. After how many minutes the length of the first candle is 3 times that of the other? (b) 120
16. A person walks 25.0° north of east for 3.18 km. How
(a) 2T
11. There are two candles of same length and same size.
(a) 90
PHYSICS
(d) 34
A
B
(I) move A to the right along Xaxis closer to B (II) move A to the left along Xaxis away from B
KVPY Question Paper 2016 Stream : SA
50 (III) as viewed from above, rotate A clockwise about Y axis (IV) as viewed from above, rotate A anticlockwise about Y axis Which of the actions will induce a current in A only in the direction shown? (a) Only (I) (c) Only (I) and (IV)
(b) Only (II) (d) Only (II) and (III)
22. A rigid ball rolls without slipping on a surface shown below:
nucleus. The lead nucleus has A =206. The electrostatic force between two protons in this nucleus is approximately (a) 102 N
(b) 107 N
(c) 1012 N
(d) 1017 N
26. A hollow lens is made of thin glass and in the shape of a double concave lens. It can be filled with air, water of refractive index 1.33 or CS2 of refractive index 1.6. It will act as a diverging lens, if it is (a) filled with air and immersed in water (b) filled with water and immersed in CS2 (c) filled with air and immersed in CS2 (d) filled with CS2 and immersed in water
27. A stone thrown down with a speed u takes a time t1 Which one of the following is the most likely representation of the distance travelled by the ball versus time graph? Distance
Distance (a)
Distance (c)
(b) Time
Time
Distance
(a)
1 gt1 t2 2
(b) g /8 (t1 + t2 )2
(c) g /8(t1 − t2 )2
(d)
Time
to reach the ground, while another stone thrown upwards from the same point with the same speed takes time t2. The maximum height the second stone reaches from the ground is
Time
23. In an experiment, set up A consists of two parallel wires which carry currents in opposite directions as shown in the figure. A second set up B is identical to set up A, except that there is a metal plate between the wires.
(d)
1 2 gt2 2
28. An electric field due to a positively charged long straight wire at a distance r from it is proportional to r −1 in magnitude. Two electrons are orbiting such a long straight wire in circular orbits of radii 1 A and 2A . The ratio of their respective time periods is (a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 4 : 1
29. Two particles of identical mass are moving in circular orbits under a potential given by V (r ) = Kr − n , where K is a constant. If the radii of their orbits are r1. r2 and their speeds are v1 ⋅ v2, respectively. Then,
(a) v12r1n = v22r2n Set up A
Set up B
Let FA and FB be the magnitude of the force between the two wires in setup A and setup B, respectively. (b) FA < FB (d) FA > FB = 0
(a) FA > FB ≠ 0 (c) FA = FB ≠ 0
24. In the circuit, wire 1 is of negligible resistance. Then, R2
R1
+ – ε1
Wire 1
ε2
(d) no current will flow through wire 1
25. The radius of a nucleus is given by r0 A1/ 3 , where r0 = 13 . × 10
(d) v12r12 − n = v22r22 − n
30. Mercury is often used in clinical thermometers. Which one of the following properties of mercury is not a reason for this ? (a) The coefficient of the thermal expansion is large (b) It is shiny (c) It is a liquid at room temperature (d) It has high density
CHEMISTRY + –
(a) current will flow through wire 1, if ε1 ≠ ε2 ε ε (b) current will flow through wire 1, if 1 ≠ 2 R1 R2 ε +ε ε −ε (c) current will flow through wire 1, if 1 2 ≠ 1 2 (R1 + R2 ) (R1 − R2 )
−15
(c) v12r1 = v22r2
(b) v12r1− n = v22r2− n
m and A is the mass number of the
31. One mole of one of the sodium salts listed below, having carbon content close to 14.3% produces 1 mole of carbon dioxide upon heating (atomic mass of Na = 23, H = 1, C = 12, O = 16). The salt is (a) C2H5COONa (c) HCOONa
(b) NaHCO3 (d) CH3COONa
32. Among formic acid, acetic acid, propanoic acid and phenol, the strongest acid in water is (a) formic acid (c) propanoic acid
(b) acetic acid (d) phenol
51
KVPY Question Paper 2017 Stream : SA 33. According to Graham’s law, the rate of diffusion of CO, O2 , N2 and CO2 follows the order
44. The electronic configuration, which obeys Hund’s rule for the ground state of carbon atom is
(a) CO == N2 > O2 > CO2
(b) CO == N2 > CO2 > O2
2p
(c) O2 > CO == N2 > CO2
(a) Energy
(d) CO2 > O2 > CO == N2
34. The major product formed when 2butene is reacted
2p (b) Energy
2s
2s
1s
1s
with O3 followed by treatment with Zn/H2O is
35. The IUPAC name for the following compound is CH3 — CH2— CH2— CH2— C— CH2— CH2— CH3  CH2 (a) 2propylhex1ene (b) 2butylpent1ene (c) 2propyl2butylethene (d) Propyl1butylethene
36. The major products obtained in the reaction of oxalic
2p (c) Energy
(b) CO, SO2 , H2O
(c) H2S, CO, H2O
(d) HCOOH, H2S, CO
(a)
(c)
(a) only water (c) only alkalis
Photoelectric current
38. The oxidation number of sulphur is −4 in
39. Al2O3 reacts with
(b)
Intensity of radiation
(b) 1.832 (d) 1.088 (b) CS2 (d) Na 2SO3
1s
for a monochromatic source of frequency above the threshold frequency is
Li = 7). The amount of CO2 (in g) consumed by 1 g of LiOH is closest to
(a) H2S (c) Na 2SO4
2s
45. The graph that depicts Einstein’s photoelectric effect
37. LiOH reacts with CO2 to form Li2CO3 (atomic mass of (a) 0.916 (c) 0.544
2p (d) Energy
1s
acid with conc. H2SO4 upon heating are
(a) CO, CO2 , H2O
2s
Photoelectric current
(d) CH2 == CH2
(b) only acids (d) both acids and alkalis
Intensity of radiation
(d)
Photoelectric current
(b) CH3 CHO
(c) CH3 CH2OH
Photoelectric current
(a) CH3 COOH
Intensity of radiation
Intensity of radiation
40. The major product formed in the oxidation of acetylene by alk. KMnO4 is
(a) ethanol (c) formic acid
BIOLOGY
(b) acetic acid (d) oxalic acid
41. In a closed vessel, an ideal gas at 1 atm is heated from 27°C to 327°C. The final pressure of the gas will approximately be (a) 3 atm (c) 2 atm
(b) N (d) Be
43. A redox reaction among the following is (i) CdCl2 + 2KOH → Cd(OH)2 + 2KCl (ii) BaCl2 + K 2SO4 → BaSO4 + 2KCl (iii) CaCO3 → CaO + CO2 (c) (iii)
(a) 22 nm (c) 2.2 m
(b) 0.22 mm (d) 22 m
consists of
largest atomic radius is
(iv) 2Ca + O2 → 2CaO (a) (i) (b) (ii)
66 . × 109 bp?
47. The Diphtheria, Pertussis, Tetanus (DPT) vaccine
(b) 0.5 atm (d) 12 atm
42. Among the elements Li, N, C and Be, one with the (a) Li (c) C
46. What is the length of human DNA containing
(d) (iv)
(a) live attenuated strains of diphtheria, pertussis, Tetanus (b) toxoid of diphtheria, tetanus and heatkilled whole cells of Pertussis (c) whole cell lysate of diphtheria, pertussis, tetanus (d) heatkilled strains of diphtheria, pertussis, tetanus
48. Which of the following is not an enzyme? (a) Lipase (c) Trypsin
(b) Amylase (d) Bilirubin
49. The pH of the avian blood is maintained by (a) HCO3−
(b) H2PO4−
(c) CH3 COO− (d) Cl −
KVPY Question Paper 2016 Stream : SA
52 50. Podocyte layer that provides outer lining to the surface of glomerular capillaries are found in
(a) Bowman’s capsule (c) renal artery
(b) loop of Henle (d) ureter
cytosine content?
pellagra?
53. In Escherichia coli, how many codons code for the standard amino acids?
(b) 60 (d) 20
58. ‘Chipko Movement’ in the year 1974 in Garhwal (a) protecting tigers (b) preventing soil erosion by planting trees (c) preventing pollution by closing down industries (d) hugging trees to prevent the contractors from felling them
59. Which of the following amino acids is not involved in
54. Bombyx mori (silkworm) belongs to the order
gluconeogenesis? (a) Alanine (c) Glutamate
(b) Diptera (d) Coleoptera
55. The source of mammalian hormone ‘relaxin’ is (a) ovary (c) intestine
(b) 45 (d) 47
Himalayas involved
(b) Nicotine (d) Tryptophan
(a) Lepidoptera (c) Hymenoptera
57. What is the number of chromosomes in an individual (a) 44 (c) 46
52. Which one of the following is incapable of curing
(a) 64 (c) 61
(b) Bat (d) Frog
with Turner’s syndrome?
(b) 30% (d) 80%
(a) Niacine (c) Nicotinamide
link between reptiles and mammals? (a) Platypus (c) Armadillo
51. If a dsDNA has 20% adenine, what would be its (a) 20% (c) 40%
56. Which one of the following animals is a connecting
(b) Lysine (d) Arginine
60. Which of the following entities causes syphilis? (a) Treponema pallidum (c) HIV
(b) stomach (d) pancreas
(b) Neisseria gonorrhoeae (d) HepatitisB
PARTII (2 Marks Questions) MATHEMATICS
64. Let S1 be the sum of areas of the squares whose sides
61. Suppose a is a positive real number such that 5
3
a − a + a = 2 . Then, (a) a 6 < 2 (c) 3 < a 6 < 4
(b) 2 < a 6 < 3 (d) 4 ≤ a 6
are parallel to coordinate axes. Let S2 be the sum of areas of the slanted squares as shown in the figure. S Then, 1 is equal to S2
62. Consider the quadratic equation nx2 + 7 nx + n = 0, where n is a positive integer. Which of the following statements are necessarily correct? I. For any n, the roots are distinct. II. There are infinitely many values of n for which both roots are real. III. The product of the roots is necessarily an integer. (a) III only (c) II and III
(b) I and III (d) I, II and III
63. Consider a semicircle of radius 1 unit constructed on the diameter AB and let O be its centre. Let C be a point on AO such thatAC : CO = 2 : 1. Draw CD perpendicular to AO with D on the semicircle. Draw OE perpendicular to AD with E on AD. Let OE and CD intersect at H. Then, DH equals (a)
1 5
(b)
1 3
(c)
1 2
(d)
5 −1 2
(a) 2
(b) 2
(c) 1
(d)
1 2
65. If a 3digit number is randomly chosen. What is the probability that either the number itself or some permutation of the number (which is a 3digit number) is divisible by 4 and 5? (a)
1 45
(b)
29 180
(c)
11 60
(d)
1 4
53
KVPY Question Paper 2017 Stream : SA 70. A Vshaped rigid body has two identical uniform
PHYSICS 66. Which one of the following four graphs best depict the variation with x of the moment of inertia I of a uniform triangular lamina about an axis parallel to its base at a distance x from it?
arms. What must be the angle between the two arms, so that when the body is hung from one end the other arm is horizontal? (a) cos−1 (1/3) (c) cos−1 (1/4)
(b) cos−1 (1/2) (d) cos−1 (1/6)
CHEMISTRY h
71. In the following reaction, X, Y and Z are
x
NO2
l
l
+ X
(a)
CH3
CH3 Z
Y
(b)
l
(a) X = CH3Cl; Y = Anhydrous AlCl 3; Z = HNO3 + H2SO4 (b) X = CH3COCl; Y = Anhydrous AlCl 3; Z = HNO3 + H2SO4 (c) X = CH3Cl; Y = Conc. H2SO4; Z = HNO3 + H2SO4 (d) X = CH3Cl; Y = Dil. H2SO4; Z = HNO3
(d)
72. 2,3dibromobutane can be converted to 2butyne in a
x
h
h
l (c)
two steps reaction using
x
h
x
h
x
67. A rectangular block is composed of three different
glass prisms (with refractive indices µ1 , µ 2 and µ3 ) as shown in the figure below. A ray of light incident normal to the left face emerges normal to the right face. Then, the refractive indices are related by µ1 45°
µ2
(d) µ 22 + µ 32 = 2 µ 12
68. A uniform metal plate shaped like a triangle ABC has a mass of 540 g. The length of the sides AB, BC and CA are 3 cm, 5 cm and 4 cm, respectively. The plate is pivoted freely about the point A. What mass must be added to a vertex, so that the plate can hang with the long edge horizontal? (b) 540 g at C (d) 540 g at B
69. A 20 g bullet whose specific heat is 5000 J kg°C and moving at 2000 m/s plunges into a 1.0 kg block of wax whose specific heat is 3000 J kg°C. Both bullet and wax are at 25°C and assume that (i) the bullet comes to rest in the wax and (ii) all its kinetic energy goes into heating the wax. Thermal temperature of the wax (in °C) is close to (c) 37.9
O3 ( g) → 3 / 2 O2 ( g); ∆H = − 1423 . kJ/mol ∆H = + 4950 . kJ/mol
NO( g) + O( g) → NO2 ( g)
45°
(c) µ 12 + µ 32 = 2µ 22
(b) 31.5
∆H = − 1989 . kJ/mol
The enthalpy change (∆H ) for the following reaction is
(b) µ 12 + µ 22 =µ 32
(a) 28.1
73. Given, NO( g) + O3 ( g) → NO2 ( g) + O2 ( g);
O2 ( g) → 2O ( g);
µ3
(a) µ 12 + µ 22 = 2µ 32
(a) 140 g at C (c) 140 g at B
(a) (i) HCl and (ii) NaH (b) (i) alc.KOH and (ii) NaNH2 (c) (i) Na and (ii) NaOH (d) (i) Br2 and (ii) NaH
(d) 42.1
(a) −3041 . kJ/mol (b) +304.1 kJ/mol (c) −4031 . kJ/mol (d) +403.1 kJ/mol
74. A 1.85 g sample of an arsenic containing pesticide
was chemically converted to As O34− (atomic mass of As = 74.9) and titrated with Pb2+ to form Pb3 (AsO4 )2 . If 20 mL of 0.1 M Pb2+ is required to reach the equivalence point, the mass percentage of arsenic in the pesticide sample is closest to
(a) 8.1 (c) 5.4
(b) 2.3 (d) 3.6
75. When treated with conc. HCl. MnO2 yields a gas (X) which further reacts with Ca(OH)2 to generate a white solid (Y). The solid Y reacts with dil. HCl to produce the same gas X. The solid Y is
(a) CaO (c) Ca(OCl)Cl
(b) CaCl 2 (d) CaCO3
76. The atmospheric pressure is 760 mm Hg at the sea
P.
level. Which of the following ranges is nearest to the partial pressure of CO2 in mm Hg?
(a) 0.300.31 (c) 3.03.1
Q.
(b) 0.600.61 (d) 6.06.1
Reaction rate
BIOLOGY
Reaction rate
KVPY Question Paper 2016 Stream : SA
54
Temperature
Temperature
(a) 1215 (c) 540
R.
Temperature
(b) 405 (d) 135
Temperature
(a) P and P (c) P and R
78. Match the different types of heart given in Column I with organisms given in Column II. Choose the correct combination. Column I
S.
Reaction rate
flowers to a pure bred short plant having blue flowers. He obtained 202 F1 progeny and found that they are all tall having white flowers. Upon selfing these F1 plants, he obtained a progeny of 2160 plants. Approximately, how many of these are likely to be short and having blue flowers?
Reaction rate
77. A breeder crossed a pure bred tall plant having white
(b) P and S (d) R and R
80. Match the enzymes in Column I with the reactions in Column II. Select the correct combination. Column I
Column II
Column II
P. Neurogenic heart
i. Human
P. Hydrolase i.
Interconversion of optical isomers
Q. Bronchial heart
ii. King crab
Q. Lyase
R. Pulmonary heart
iii. Shark
Oxidation and reduction of two substrates
ii.
R. Isomerase iii. Joining of two compounds
(a) Pii, Qiii, Ri (b) Piii, Qii, Ri (c) Pi, Qiii, Rii (d) Pii, Qi, Riii
S. Ligase
iv. Removal of a chemical group from a substance v.
79. Given below are the four schematics that describe the dependence of the rate of an enzymatic reaction on temperature. Which of the following combinations is true for thermophilic and psychrophilic organisms?
Transfer of a chemical group from one substrate to another
(a) Piv, Qii, Riii, Si (c) Piv, Qi, Riii, Sv
(b) Pv, Qiv, Ri, Siii (d) Pi, Qiv, Rv, Sii
Answers PARTI 1
(c)
2
(c)
3
(b)
4
(d)
5
(a)
6
(b)
7
(b)
8
(c)
9
(c)
10
(b)
11
(d)
12
(c)
13
(a)
14
(c)
15
(d)
16
(d)
17
(c)
18
(c)
19
(c)
20
(b)
21
(a)
22
(d)
23
(c)
24
(d)
25
(a)
26
(d)
27
(b)
28
(b)
29
(a)
30
(d)
31
(b)
32
(a)
33
(a)
34
(b)
35
(a)
36
(a)
37
(a)
38
(*)
39
(d)
40
(d)
41
(c)
42
(a)
43
(d)
44
(a)
45
(c)
46
(c)
47
(b)
48
(d)
49
(a)
50
(a)
51
(b)
52
(b)
53
(c)
54
(a)
55
(a)
56
(a)
57
(b)
58
(d)
59
(b)
60
(a)
PARTII 61
(c)
62
(b)
63
(c)
64
(a)
65
(b)
66
(a)
67
(c)
68
(c)
69
(c)
70
(a)
71
(a)
72
(b)
73
(a)
74
(c)
75
(c)
76
(a)
77
(d)
78
(a)
79
(d)
80
(b)
* No option is correct.
55
KVPY Question Paper 2017 Stream : SA
Solutions 1. (c) We have, p(x) = ax2 + bx + c, where a , b, c are in AP and a , b, c are positive real. α , β are root of p (x) = 0, where α and β are integers. p (x) = ax2 + bx + c = 0 −b c α+β= , αβ = a a α , β are integer. −b ∴ α+β= = − λ, λ ∈ I a ⇒ b = aλ a , b, c are in AP. a+ c a+ c b= ⇒ = aλ ∴ 2 2 ⇒ c = a (2λ − 1) ∴ ax2 + aλx + a (2λ − 1) = 0 ⇒ x2 + λx + (2λ − 1) = 0 [Q a ≠ 0] 2 D = λ − 4 (2λ − 1) is a perfect square for integral roots. Q λ2 − 8λ + 4 = k 2 ⇒ (λ − 4)2 − 12 = k 2 ⇒ (λ − 4 − k ) (λ − 4 + k ) = 2 × 6 ⇒ λ − 4 − k = 2 and λ − 4 + k = 6 Q λ = 8 and k = 2 −b c + ∴ α + β + αβ = a a − aλ + a (2λ − 1) = a a (λ − 1) = a = λ − 1= 8 − 1= 7
(iii) 10t + 3 a 2 + b = 10, ab = 3 a = 3, b = 1 it is possible (iv) 6t + 5 a 2 + b = 6, ab = 5 a = 1, b = 5 it is also possible Hence, option (b) is correct.
…(i)
1 × BE × h 2 1 = × 5× h 2 From Eqs. (i) and (ii), 1 1 × 3× 4= × 5× h 2 2 ∆BCE =
(1 + a + b)2 = 3 (1 + a 2 + b2 ) 1 + a + b2 + 2a + 2b + 2ab 2
⇒
= 3 + 3a 2 + 3b2
…(ii)
h = 2.4
7. (b) Given,
⇒ 2a 2 + 2b2 − 2a − 2b − 2ab + 2 = 0 (a 2 − 2a + 1) + (b2 − 2b + 1)
⇒
1 × 3× 4 2
Also, area of
4. (d) Given,
+ (a 2 + b2 − 2ab) = 0
∠ABC = 130° ∠CDE = 110° ABCE is a cyclic quadrilateral.
⇒ (a − 1)2 + (b − 1)2 + (a − b)2 = 0 ∴ a − 1 = 0, b − 1 = 0, a − b = 0 ⇒ a = 1, b = 1, a = b ∴ a= b=1 Exactly one pair.
A 70° E
50°
130°
B
5. (a) We have, a1 , a2 , a3 , ..., a100 be nonzero real number and a1 + a2 + a3 + ... + a100 = 0 ai ⋅ 2ai > ai and ai ⋅ 2− ai < ai ∴
100
∑ a1 ⋅ 2
ai
>
i =1 100
100
100
i =1
i =1
∑ ai and ∑ a1 ⋅ 2
− ai
0 and ∑ a1 ⋅ 2− ai < 0
2. (c) We have,
⇒ t3 = a (at + b) + bt ⇒ t3 = a 2t + bt + ab ⇒ t3 = (a 2 + b)t + ab (i) 4t + 3 a 2 + b = 4, ab = 3 a = 1, b = 3 it is possible (ii) 8t + 5 a 2 + b = 8, ab = 5 It is not possible
=
4
1
1
h
2 3
A
6
11
E
5
CE = 3 BC = 4 BE = 5 ∴ ∠BCE is a right angled triangle. 1 ∴Area of ∆BCE = EC × BC 2
B
A
3
∴
∴Side of ∆ ABC are AB = 5 BC = 3 AC = 4
2
B
KVPY Question Paper 2016 Stream : SA
56
A
M P1
P
B
C
12. (c) Given, a cuboid has all edges are
P2
Circumcentre of ∆ P1 PP2 is midpoint of P1 P2. AB is perpendicular bisector of PP1 and BC is perpendicular bisector of PP2. Perpendicular bisector of PP1 and PP2 intersect at B.
integers and base is square. Let the length, breadth and height of cuboid is x, x, y.
∴ B is circumcentre of ∆P1 PP2. ∴Distance between BM = AM = MC =
AC . 2
10. (b) Given, a + 2b ≤ 1 a, b are positive real number. Radius of circle C1 = ab3 Radius of circle C2 = b2 ∴Area of circle C1 = A1 = πa 2b6 and area of circle C2 = A2 = πb4 A1 πa 2b6 Now, = = a 2b2 A2 πb4 ⇒
a + 2b ≤ 1 a + 2b ≥ 2ab 2
[Q AM ≥ GM] 1 ≥ (a + 2b)2 ≥ 8ab 8ab ≤ 1 1 a 2b2 ≤ ⇒ 64 A1 1 ≤ ∴ A2 64 A 1 ∴Maximium value of 1 = A2 64 ⇒ ⇒
x
y x
Sum of all edges of cuboid = 4x + 4x + 4 y Sum of area of all faces = 2x2 + 2xy + 2xy Given, Sum of all edges of cuboid = Sum of area of all faces ∴ 4x + 4x + 4 y = 2 (x2 + xy + xy) ⇒ 4 (2x + y) = 2 (x2 + 2xy) 2 ⇒ x + 2xy − 4x − 2 y = 0 ⇒ x2 + 2x( y − 2) − 2 y = 0 ⇒ ⇒
x=
− 2 ( y − 2) ±
x = y − 2±
A1 ∩ A2 ∩ A3 ... ∩ Am = φ ∴ A1 ∩ A2 ∪ A3 ... ∪ Am = {1, 2, 3, ..., 100} LetA1 = 1A2 = 2 ....Am = M A1 , A2 , A3 …, Am are disjoint set. ∴A1 + A2... + Am= 100 1 + 2 + 3 .... + m = 100 m (m + 1) = 100 2 m2 + m − 200 = 0 b2 − 4ac −1 ± 1 + 4 ⋅ 1⋅ 200 = 2a 2⋅ 1 −1 ± 1 + 800 −1 + 801 = = 2 2 −1 + 28.30 27.30 = = = 16.65 2 2 1 + 28.30 29.30 m= = = 14.65 2 2 ∴m < 14 ∴Maximum possible of m is 13. (14th set will have same size as that of previous size) =
−b ±
14. (c) Let twodigits number be n = 10a + b Given, n = a 2 + b3 ∴
10a + b = a 2 + b3
⇒ a 2 − 10a + b3 − b = 0 ⇒ a (a − 10) + b (b + 1) (b − 1) = 0 ⇒ b (b + 1) (b − 1) = a (10 − a ) b ≠ 1if b = 1then a = 10 not possible if b = 2, a (10 − a ) = 6, no value of a b = 3, a (10 − a ) = 24, a = 4, 6 Numbers are 43 and 63. If b = 4, a (10 − a ) = 60 no value of a If b = 5, a (10 − a ) = 120 not possible ∴Numbers are 43 and 63.
15. (d) Given, f (xy) = f (x) + f ( y) f (12) = 24 ⇒ f (8) = 15 f (8) = f (2 ⋅ 2 ⋅ 2) = f (2) + f (2) + f (2) ⇒ 15 = 3f (2) ⇒ f (2) = 5 ∴ f (48) = f (12 ⋅ 2 ⋅ 2) = f (12) + f (2) + f (2) = 24 + 5 + 5 = 34 16. (d) Displacement of person is
4( y − 2)2 + 4(2 y)
N
2 2
y − 2y + 4
x is integer, when y = 2 ∴ y = 2, x = 2 Hence, sum of edges = 8x + 4 y = 16 + 8 = 24
25° W
O
A
13. (a) We have, A1 , A2 , A3 ...., Am are nonempty subsets of {1, 2, 3, …, 100} A1,A2,...,Amare distincts.
B
km
∠ABC = 90° Circumcentre of ∆ ABC is midpoint of AC i.e. M.
candles are same. Let L be the length of candles. Given, first candle burns in 5 h and second candle burns in 3 h. L In one hours length of candles are and 5 L , respectively. 3 Let after time t h the length of candles are L1 and L2. L L L1 = L − t and L2 = L − t ∴ 5 3 According to the problem, L1 = 3L2 L L ∴ L − t = 3 L − t 5 3 1 1 ⇒ 1 − t = 3 − t ⇒ t 1 − = 3 − 1 5 5 5 4t = 2 ⇒t = h ⇒ 2 5 5 ⇒ t = × 60 = 150 min 2
18
9. (c) ABC is a right angled triangle,
11. (d) We have, length and size of two
3.
∴∆ABC is formed a right angled triangle where AB is hypotenuse of triangle. We know circumradius of a right angled triangle is the half of the hypotenuse. 1 ∴Circumradius = × AB 2 1 = × 5 = 2.5 2
S
E
57
KVPY Question Paper 2017 Stream : SA From above figure, distance travelled along north direction is AB = OB sin 25° = 318 . × sin 25° = 1.34 km Distance travelled along east direction is OA = 318 . × cos 25° = 2.88 km
21. (a) Current in loop is shown
25. (a) Taking protons at dimetrically
anticlockwise.
opposite points of nucleus, A
IA
IB
B
Now, ⇒ ⇒
= 3.95 × 3.05 = 12.05 m 2 A=l×b dA dl db = + A l b ∆l ∆b + ∆A = ×A l b 0.05 0.05 = + × 12.05 3.95 3.05
≈ 0.34 So, area of floor is A = 12.05 ± 0.34 m 2.
18. (c) When car goes around track of radius R, then Time period,
T =
2πR v
Hence to induce a anticlockwise current in A, flux going into A must be increased and by bringing A closer to B, we get a anticlockwise current in A. This is in accordance with Lenz’s law.
22. (d) As ball moves from point A to point D, A
B C
v2 Centripetal acceleration, ac = R When car goes around circular track of radius 2R, then Centripetal acceleration, v′2 8v2 ac′ = = ⇒ v′ = 4v 2R R So, time period of car is 2 πR ′ 2 π (2R ) T′ = = 4v v′ 1 2 πR T = × = 2 v 2
19. (c) As primary voltage is constant, there is no change of magnetic flux of secondary coil. So, there is no induction and hence voltage across secondary is zero.
D
Velocity of ball increases rapidly in region BC. Then, from C to D, it moves with a constant velocity. So, graph is parabolic in region BC and slope of AB is less than slope of CD, as shown below. x D Straight line
C
Parabolic curve
B A
potential energy of water Received by generator per second 50 mgh = × 100 t
P2
Lead nucleus
Force of electrostatic repulsion, k (e) (e) ke2 k ⋅ e2 F= = = 2 2 /3 1/3 2 2 r (r0 A ) r0 ⋅ A Substituting values in above equation, we get . × 10−19 )2 9 × 109 × (16 F= (13 . × 10−15 )2 (206)2/3 = 0.039 × 102 N
26. (d) When medium outside a lens is denser than medium of lens, then a concave lens will acts like a convex lens and viceversa. Now, when hollow lens is filled with CS2 (η = 16 . ) and immersed in water (η = 133 . ) , its nature remains diverging as refractive index of medium of lens is more than refractive index of surrounding medium. 27. (b) For first stone,
Straight line
t=0
Note Velocity change is not very abrupt.
u
23. (c) Metal plate between wires may modify field pattern within the metal volume but number of field lines is not changed. So, force between wires is same in both cases and is nonzero. i.e. FA = FB ≠ 0.
20. (b) Power output, P = 50% of
V = 0.5 × ρ × g × h t
r
P1
17. (c) Area, A = l × b
24. (d) R1
R2
h
− h = − ut1 + h = ut1 +
or
1 (− g ) t12 2
1 2 gt1 2
...(i)
For second stone,
...(i)
Here, P = 1 × 109 W, g = 10 ms −2, ρ = 1000 kgm −3 and h = 500 m. Substituting these values in Eq. (i), we get V Volume flow rate of water = t 9 1 × 10 = = 400 m3 s −1 0.5 × 1000 × 10 × 500
i2 i1
E1
i1
i2
u t=0
E2
Current leaving the cell must be equal to current going into the cell. So, current going from first loop to second loop must be zero for any value of E or R. Hence, there is no current through the wire connecting loops.
h t2
1 2 ...(ii) gt2 2 Adding Eqs. (i) and (ii), we get 1 0 = u (t1 + t2 ) − g (t12 − t22 ) 2
− h = ut2 −
KVPY Question Paper 2016 Stream : SA
58 g (t1 − t2 ) 2 Maximum height attained by second stone is u2 H =h+ 2g 1 u2 ⇒ H = ut1 + gt12 + 2 2g ⇒
u=
Substituting for u and rearranging, we get g H = (t1 + t2 )2 8 28. (b) Electron revolve around wire due to its electrostatic force of attraction. As field E ∝ r −1 ⇒ E = kr −1 +
r
+ + +
So, force on electron is F = eE = ker −1 This force is necessary centripetal force. So, mv2 ke = r r ke v= ⇒ m As velocity of electron is independent of radius of paths, ⇒ v1 = v2 Now, ratio of time periods of rotation are 2 πr1 T1 v1 r1 1Å 1 = = = = T2 2 πr2 r2 2 Å 2 v2
29. (a) Given, Potential, V = Kr − n So, magnitude of field, dV d =− (Kr − n ) E=− dr dr nK E = n+1 ⇒ r mnK Hence, force, F = mE = n + 1 r As particles are rotating in circular path, mv2 mnK = n+1 F= r r ⇒ So,
v2r n = nK = constant v12r1n = v22r2n
30. (d) Fluid used in a thermometer must be easily visible, expands uniformly and significantly and it must be a liquid at room temperature. So, density of mercury is not a feature for selecting mercury in clinical thermometers. 31. (b) NaHCO3 which is a sodium salt only produces carbon dioxide on heating while the other salts produce ions on heating. Also, its carbon content which is close to 14.3%. ∆
2NaHCO3 → Na 2CO3 + H2O + CO2 Molecular mass of NaHCO3 = 23 + 1 + 12 + 16 × 3 = 84 12 % of C = × 100 = 14.28% 84
O
The substituents attached to benzoic acid having + I effects tends to decrease its acidity. • Carboxylic acids are more acidic than phenol as they are more resonance stabilised. Among the three carboxylic acids given in the options, the + I effect of CH3 group intensifies the negative charge on the carboxylate ion thereby making acetate ion less stable than formate ion. As a result, the release of H+ ion from acetic acid will become more difficult as compared to formic acid. Hence, formic acid is a stronger acid than acetic acid. O CH3
C
O
Acetate ion
–
O C
H
–
O
Formate ion
Further since + I effect of alkyl groups increases in the order CH3 —< CH3 CH2— the relative acid strength will decrease in the same order, i.e.,
C O
O –
Carboxylate ion
–
C O (More resonance stabilised)
O
–
Phenoxide ion
O
–
O –
Thus, the order of acidic strength of given compounds is OH HCOH>CH3COOH >CH3CH2COOH> Propanoic acid
Acetic acid
O
Phenol
Formic acid
Hence, formic acid is a strongest acid.
33. (a) According to Graham’s law, rate of diffusion is inversely proportional to the square root of molar mass, i.e. 1 r∝ M Thus, rate of diffusion decrease with increase in molecular weight. Therefore, the order of rate of diffusion will be CO == N2 > O2 > CO2 (28 g) (28 g) (32 g) (44 g)
34. (b) The major product, formed when 2butene is reacted with O3 followed by treatment with Zn / H2O is acetaldehyde. This reaction is known as reductive ozonolysis. CH3CH
CH
CH3 + O3
2 butene
O CH3CH
HC OH > CH3 COOH > CH3 CH2COOH  O Now between carboxylic acids and phenol, carboxylic acids are stronger acids than phenol because carboxylate ion is more resonance stabilised. This is because the negative charge on the carboxylate ion is delocalised over two electronegative oxygen atoms. O
O –
32. (a) Key Idea
•
–
O
CH
CH3
O
Ozonide Zn/H2O
2CH3
C
H
O Acetaldehyde
35. (a) The IUPAC name of the following is 6
5
4
3
2
CH3 CH2 CH2 CH2 C CH2 CH2CH3  1CH2 2 propylhex1ene
59
KVPY Question Paper 2017 Stream : SA 41. (c) According to ideal gas equation,
the reaction of oxalic acid with conc. H2SO4 upon heating are carbon monoxide, carbon dioxide and water. COOH ∆ → CO + CO2 + H2O  H SO COOH 2 4
pV = nRT nRT p= V At constant volume and number of moles p ∝T p1 T1 = ∴ p2 T2
Oxalic acid
37. (a) 2LiOH + CO2 → Li 2CO3 + H2O 1 moles 24 2 moles of LiOH reacts with 1 mole of CO2 to form 1 mole of Li 2CO3 and H2O. 1 ∴ Number of moles of CO2 = 24 × 2 1 moles = 48 1 mole of CO2 = 44 g 1 1 moles of CO2 = × 44 = 0.916 g 48 48
T1 = 27°C = 27 + 273 = 300K T2 = 327°C = 327 + 273 = 600K 1 300 = p2 600
Number of moles of LiOH =
38. (*) The oxidation number of S in given compounds are as follows : (i) H2S Let the oxidation state of S be x ∴ 2+ x= 0 x= −2 (ii) CS2 4 + 2x = 0 ⇒ x = − 2 (iii) Na 2SO4 2(+1) + x + 4(−2) = 0 2+ x− 8= 0 x=+6 (iv) Na 2SO3 2(+1) + x + 3(−2) = 0 2+ x− 6= 0 x=+4 No, option is correct.
Acetylene
CH
O
Glyoxal
43. (d) A redox reaction is one in which the oxidation and reduction reactions occur simultaneously. (i) CdCl 2 + 2KOH → Cd(OH)2 + 2KCl This is an example of double displacement reaction. (ii) BaCl 2 + K2SO4 → BaSO4 + 2KCl This is a double displacement reaction.
47. (b) DPT is a class of combination vaccines against three infectious diseases in humans, i.e. diphtheria, pertussis (whooping cough) and tetanus. The vaccine components include diphtheria and tetanus toxoids and killed whole cells of the bacterium that causes pertussis.
0
+2 –2
As both oxidation and reduction reaction occurs simultaneously, so it is a redox reaction.
44. (a) According to Hund’s rule, ‘pairing of electrons in the orbitals belonging to same subshell does not take place until each orbital belonging to that subshell has got one electron each, i.e. it is singly occupied. The electronic configuration, which obeys Hund’s rule for ground state of carbon is 1s2 2s2 2 p 2.
2[O] Alk. KMnO4
2p
Oxalic acid (Major product)
48. (d) Bilirubin is an orange yellow pigment formed in the liver by the breakdown of haemoglobin and excreted in bile. It is not an enzyme. Other options like lipase, amylase and trypsin are lipid digesting, starch digesting and endopeptidase enzymes.
49. (a) The pH of the avian blood is
Oxidation
COOH
46. (c) The distance between 2 nucleotides/nitrogen bases is 0.34 × 10−9 m or 3.4 Å. Therefore the length of human DNA containing 6.6 × 109 bp would be = 0.34 × 10−9 m × 6.6 × 109 bp = 2.244 m or 2.2 m
(iv) 2Ca + O2 → 2CaO
COOH
Intensity of radiation
and Be belong to same period i.e. 2nd period, so on moving from left to right in a period the atomic radius decreases because the effective nuclear charge increases. Thus, Li has the largest atomic radius among them all.
Reduction
oxidation of acetylene by alk. KMnO4 is oxalic acid. O
42. (a) As the given elements Li, N, C
0
40. (d) The major product formed in the
CH
p2 = 2 atm
(iii) CaCO3 → CaO + CO2
(those oxides which show both the properties of acids and bases), so it can react both with acids and alkalis, e.g. Al 2O3 + HCl → AlCl3 + H2O Al 2O3 + NaOH + H2O → Na[Al(OH)4 ]
Alk. KMnO4
⇒
Reduction
39. (d) Al 2O3 is an amphoteric oxide
CH +2[O] CH
∴
45. (c) According to photoelectric effect, the number of electrons ejected is proportional to the intensity of radiation. Thus on increasing the intensity of radiation, the value of photoelectric current also increases, i.e. photoelectric current ∝ intensity of radiation. Hence, its graph would be linear. Photoelectric current
36. (a) The major products obtained in
2s 1s
maintained by bicarbonate (HCO3− ) ions. A variety of buffering systems exist in the body of birds that helps to maintain the pH between 7.35 and 7.45. Since the bicarbonate (HCO3− ) ion is a base, it helps neutralise the acid in the blood and increases the pH above 7.
50. (a) Podocytes are cells of squamous epithelium of Bowman capsule of nephron. The Bowman’s capsule filters the blood, retaining large molecules such as proteins while smaller molecules such as water, salts and sugars are filtered as the first step in the formation of urine. 51. (b) If dsDNA has 20% adenine, then according to the Chargaff’s rule, it would have 20% thymine. The remaining 60% represents both G + C. Since guanine and cytosine are always present in equal numbers, the percentage of cytosine molecule is 30%.
KVPY Question Paper 2016 Stream : SA
60 52. (b) Pellagra cannot be cured by
60. (a) Causative agent of syphilis is
nicotine. Instead, pellagra can be cured by giving niacine, nicotinamide and tryptophan. Pellagra is a disease caused by a lack of the vitamin niacin (vitaminB3 ) which includes both nicotinic acid and nicotinamide and its precursors, i.e. the amino acid tryptophan. The main symptoms of pellagra are dermatitis, dementia and diarrhoea.
Treponema pallidum. Syphilis is a bacterial infection usually spreads by sexual contact that starts as a painless sore. Neisseria gonorrhoeae causes gonorrhoea. HIV causes AIDS. HepatitisB virus causes hepatitis, i.e. a liver infection.
between reptiles and mammals. They have few mammalian characters such as hair, mammary glands, diaphragm whereas it lays eggs with yolk and egg shell similar to reptiles.
57. (b) The people with Turner’s
63. (c) Given,
the ovary and the placenta with important effects in the female reproductive system and during pregnancy. In preparation for childbirth, it relaxes the ligaments in the pelvis and softens and widens the cervix.
56. (a) Platypus is a connecting link
syndrome have 44 + XO chromosomes, so there are a total of 45 chromosomes only in each cell. Such persons are sterile females who have rudimentary ovaries, undeveloped breasts, small uterus, short stature and abnormal intelligence.
58. (d) ‘Chipko Movement’ was headed by social activist Sunder Lal Bahuguna in Uttarakhand to save trees from felling. The movement got its name due to people’s action of hugging trees in order to prevent them from cutting down by state forest contractors.
59. (b) Lysine is not involved in gluconeogenesis. Gluconeogenesis is the biosynthesis of new glucose from certain noncarbohydrate carbon substrates like amino acids, etc. Out of the 20 amino acids, 18 are glucogenic (i.e. can be converted to glucose upon gluconeogenesis) while the remaining two amino acids, i.e. lysine and leucine are purely ketogenic (i.e. can be degraded into acetylCoA).
2 3 2 = 3 3 In ∆DEH ~ ∆OEA 1/ 3 DH DE DH = = ⇒ 2/ 3 1 OE OA =
⇒ DH =
1 2
Q DE = 1 AD 2 2 1 2 OE = OA − AD 2
64. (a) Here,
62. (b) Given, nx2 + 7 n x + n = 0 D = 49n − 4n 2 = n (49 − 4n ) D ≠ 0 ; ∴∀ n ∈ I + ∴ Roots are distinct. For roots are real D ≥ 0 49 ∴ n (49 − 4n ) ≥ 0 ⇒ n ≤ 4 So, n ∈ {1, 2, 3, 4, ..., 12} So, x have finite value. n Product of roots is = 1 n ∴Products of root is necessarily integer. Hence, option (b) is correct.
55. (a) Relaxin hormone is produced by
8 4 + 9 9
AD = CD 2 + AC 2 =
a
54. (a) The order of silkworm (Bombyx mori) is Lepidoptera. It is the order of insects that includes butterflies and moths. About 1,80,000 species of the Lepidoptera are described till now.
In ∆ACD,
1 2 2 = 9 3
2
64 codons and out of which 3 codons are stop or termination codons, i.e. UAA, UAG and UGA which do not code for any amino acids. Therefore, 61 codons code for the standard 20 amino acids.
⇒ a5 − a3 + a − 2 = 0 Let f (a ) = a5 − a3 + a − 2 f ′ (a ) = 5a 4 − 3a 2 + 1 f ′(a ) > 0, ∀ a ∈ R ∴ a5 − a3 + a − 2 = 0 has only one roots. for a 6 = 3 ⇒ a = (3)1/ 6 = 12 . [by calculation] f (41/ 6 ) > 0 and at a 6 = 4, a = (4)1/ 6 So one root lies in (3, 4). ∴ 3 < a6 < 4
CD = OD 2 − OC 2 = 1 −
2√
53. (c) In all living organisms, there are
61. (c) Given, a5 − a3 + a = 2
In ∆OCD,
a/2 a 2
2
S1 = a 2 +
S2 = 1 H C 1 O 3
AB is diameter of circle. O is centre of circle. 1 ∴ OA = OB = AB = 1 2 C is a point on AO such that AC 2 = ⇒ AC = 2OC OC 1 CD is perpendicular to AO. ∴ OD = OA radius of circle OE is perpendicular on AD. ∴ AOD is isosceles triangle. ∴ E is midpoint of AD. 1 2 ∴ OA = 1, OC = , AC = , OD = 1 3 3
B
2
2
a a + ... S2 = + 2 2 2
E
2 3
a/2
√2
a a S1 = a 2 + + + ... 4 2
D
A
a
∴
4a 2 a2 a2 a2 = + + ... = 1 3 4 16 1− 4
a 2 / 2 4a 2 a2 a2 a2 = + + + ... = 1 6 2 8 32 1− 4
4a 2 S1 3 =2 = S2 4a 2 / 6
65. (b)The 3digit number which is divisible by 4 and 5 both. i.e. last digits are 00, 20, 40, 60, 80 Now ending with 00 are (100, 200, 300 … 900) = 9 If digit repeat other than 0′ then they are (220, 440, 660, 880) but 220 number can be permuted according to condition as (220, 202). Similarly, for 440 as (440, 404), 660 and 880, so there are 8 favourable cases. If the number have no digit repeated like 120, 120 can be permuted in 4 ways. So, such number are 8 × 4 × 4 = 128
61
KVPY Question Paper 2017 Stream : SA Total favourable cases = 9 + 8 + 128 = 145 145 29 Required probability = = 900 180
66. (a) By parallel axes theorem, moment of inertia of triangular lamina about a parallel axes, which passes below its centre of mass is CM
4
x
C
Clearly, I versus x is a parabolic graph. Also, I first reduces axis of rotation comes closer to centre of mass and then it again increases. So, correct variation of I with x is I
h
x
x=h/3
67. (c) Ray diagram of ray through the A α
1i =
45
°
composition of prisms will be µ1
45°
θ
µ3 µ2 α–θ
45°
45°
B
C
By Snell’s law on surface AB and AC, we have ...(i) µ 1 sin 45° = µ 2 sin θ and ...(ii) µ 3 sin 45° = µ 2 cos θ As α − θ = 45°, from figure Squaring and adding Eqs (i) and (ii), µ 12 µ 32 we get + = µ 22 2 2
68. (c) Given situation is A
G 53° B D
E
y
H
2
When axis of rotation of lamina, passes above its centre of mass and its moment of inertia is 2 h I = ICM + M x − 3
ICM
angle between them is θ.
3 G
h I = ICM + M − x 3
C
A
Substituting values in above equation, we get 20 × 10−3 × (2000)2 ∆T = 2(20 × 10−3 × 5000 + 1 × 3000) 400 = 12.9 ⇒ Tf − Ti = 12.9 ⇒ ∆T = 31 or Tf = 25 + 12.9 = 37.9° C
70. (a) Let length of each of rod is l and
2h 3 h –x 3
Weight of lamina acts through its centroid G to prevent tilting of lamina, let a mass m1 is added at vertex B. From A, perpendicular AE is dropped on BC. AD is medium and G is centroid of ∆ABC. Now, consider ∆ABC and ∆EBA.
m1g
5/2
D
E
x
B
∆ABC ~ ∆EBA AB 2 9 AB BC ⇒ EB = x = = ⇒ EB = x = BC 5 EB AB So, DE = BD − EB 5 9 25 − 18 7 cm = − = = 2 5 10 10 Now, consider ∆ADE, G is centroid of ∆ABC. AG 2 2 So, = or AG = AD GD 1 3 Also, GH is parallel to DE. AG GH So, = AD DE 2 AD × DE AG × DE 3 = ⇒ GH = AD AD 2 7 14 7 cm = × = = 3 10 30 15 For BC to remain horizontal, torque of m1 g about A must be balanced by torque of mg about A. ⇒ mg × GH = m1 g × BE 7 9 = m1 × 540 × ⇒ 15 5 540 × 7 × 5 ⇒ m1 = = 140 g 15 × 9 So, mass of 140 g must be added to vertex B, so that BC remains horizontal.
69. (c) As kinetic energy of bullet is used up in heating and melting the wax. By energy conservation, we have 1 mb vb2 = mw cw (∆Tw ) + mb cb (∆Tb ) 2 As both bullet and wax initially are at same temperature (Ti = 25° C). So, ∆Tw = ∆Tb = ∆T (say) 1 Then, mb vb2 = (mw cw + mb cb ) ∆T 2 mb vb2 or ∆T = 2(mw cw + mb cb )
θ
l
P Rod 1 A θ C mg
B
Rod 2
D mg
Let the lower rod is horizontal and upper rod makes θ angle with horizontal. Weights of rods acts vertically downwards from their centres A and B as shown in above figure. Now, perpendicular distance of weight acting through point A from point D is l CD = l cos θ − cos θ 2 l CD = cos θ 2 and perpendicular distance of weight acting through B from point D is l l BD = − l cos θ = (1 − 2 cos θ) 2 2 At equilibrium torque of these two weights about D must balance each other, l l i.e. mg × cos θ = mg × (1 − 2 cos θ) 2 2 3 1 cos θ = ⇒ 2 2 1 cos θ = ⇒ 3 1 or θ = cos−1 3
KVPY Question Paper 2016 Stream : SA
62
74. (c) 3Pb2+ + 2AsO4 → Pb3 (AsO4 )2
71. (a) For the given reaction, CH3
FriedelCraft's reaction
Benzene
Toluene
Normality = Molarity × Volume
CH3
N
NO2 HNO3/H2SO4;(Z)
∴
Nitration
2nitro toluene
When benzene reacts with CH3Cl in presence of anhyd.AlCl3 , then toluene is formed. This reaction is known as FriedelCraft’s reaction. The formed toluene then undergoes nitration to give 2nitrotoluene.
72. (b) For the conversion of 2,3dibromobutane to 2butyne following steps can be used Step 1
H 3C
Br
Br
CH
CH
CH3
2, 3dibromobutane
Alc.KOH –HBr
Br CH3
CH
C
CH3
2bromo but2ene
In this step, dehydrohalogenation occurs where 2,3dibromobutane gets converted into 2bromobut2ene. Step 2
CH
Pb 2 +
= 01 . ×
20 = 2 × 10−3 1000
2 × 2 × 10−3 = 0.00133 3 = N 3− = 0.00133
N AsO 4 = N As
AsO
4
WAs = N As × MassAs = 0.00133 × 74.9 = 0.0996 0.0996 % of As = × 100 1.85 ≈ 5.38 ≈ 5.4%
75. (c) When treated with conc.HCl, MnO2 yields a chlorine gas (X), which further reacts with Ca(OH)2 to generate calcium oxychloride CaOCl 2(Y ), which is a white solid that then reacts with dil. HCl to produce again chlorine gas (X ). The equations can be written as MnO2 + HCl (conc.) → Cl 2 ( g ) X
Ca(OH)2 + Cl 2 → CaOCl 2 Y
CaOCl 2 + dil. HCl → Cl 2 + CaCl 2 + H2O (X)
Thus, solid Y is Ca(OCl)Cl.
Br CH3
reacts with 2 moles AsO4
∴1 mole of Pb will react with 2 = moles of AsO4 3
Anhyd. AlCl3 (Y)
+ CH3Cl (X)
3 moles of Pb
2+
C — CH3 + NaNH2 CH3
C
C
–NaBr –NH3
CH3
In this step, also dehydrohalogenation occurs where alkenyl halide on treatment with soda amide gives 2butyne.
73. (a) Given, NO( g ) + O3 ( g ) → NO2 ( g ) + O2 ( g ); ∆H1 = − 198.9 kJ/mol 3 O3 → O2 ( g ) ∆H 2 2 = − 142.3 kJ/mol O2 → 2O2 ( g ) ∆H3 = + 495.0 kJ/mol For the reaction, NO( g ) + O( g ) → NO2 ( g ) 1 ∆H = ∆H1 − ∆H 2 − ∆H3 2 1 = − 198.9 − (−142.3) − × 495 2 = − 3041 . kJ/mol
76. (a) pCO2 = 0.30 − 0.31 mm Hg in air. Air contains 0.04% of carbon dioxide. This means that in every 100 molecules of air, 0.04 will be CO2 molecules. The number of moles of carbon dioxide in 100 molecules of air will be nCO2 = 0.04 molecules × N A = 0.04 × N A The total number of moles in the sample of air will be ntotal = 100 molecules × N A = 100N A This means that mole fraction of carbon dioxide in the mixture will be 0.04N A = 0.0004 XCO2 = 100N A Carbon dioxide’s partial pressure in air will thus be pCO2 = 0.0004 × 760 mm Hg = 0.304 mm Hg ∴We can say it ranges between 0.300.31 mm Hg.
77. (d)
TTWW × ttww Tall plant Short plant with white with blue flowers flowers TtWw (Tall plant with white flower) (202 plants) Selfing Obtained 2160 plants total
According to dihybrid phenotypic ratio 9 : 3 : 3 : 1, TW — 9 Tw — 3 tW — 3 tw — 1 The total number of short and blue flowered plants is 1080 1 × 2130 = = 135 8 16
78. (a) P–ii, Q–iii, R–i – A neurogenic heart requires nervous input to contract. It is seen in crustaceans like king crab. – Bronchial hearts are myogenic accessory pumps found in coleoid cephalopods like shark that supplement the action of the main, systemic heart. – Pulmonary heart is found in humans where the portion of the circulatory system carries deoxygenated blood away from the right ventricle of the heart to the lungs and returns oxygenated blood to the left atrium and ventricle of the heart.
79. (d) Both thermophiles and psychrophiles will show same enzymatic reaction graph. Mostly proteinaceous enzymes are labile to temperature. Thermophiles live at very high temperature while psychrophiles live in the range of −20° C to +10°C. In either case, rising temperature will first raise the rate of reaction but if temperature is still raised continuously, enzymes get denatured, hence reaction rate decreases.
80. (b) P–v, Q–iv, R–i, S–iii – Hydrolases catalyse transfer of a chemical group from one substrate to another. – Lyase catalyses removal of chemical groups from a substrate. – Isomerase catalyses interconversion of optical, geometric or positional isomers. – Ligase catalyses linking together of two compounds.
63
KVPY Question Paper 2015 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2015 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS
4. In the figure given below, a rectangle of perimeter
1. Two distinct polynomials f (x) and g(x) are defined as
76 units is divided into 7 congruent rectangles.
follows: f (x) = x2 + ax + 2; g(x) = x2 + 2x + a. If the equations f (x) = 0 and g(x) = 0 have a common root, then the sum of the roots of the equation f (x) + g(x) = 0 is (a) −
1 2
(b) 0
(c)
1 2
(d) 1
2. If n is the smallest natural number such that
n + 2n + 3n + ... + 99n is a perfect square, then the number of digits of n 2 is
(a) 1
(b) 2
(c) 3
(d) more than 3
3. Let x, y, z be positive reals. Which of the following implies x = y = z ?
I. x3 + y3 + z3 = 3xyz
II. x3 + y 2z + yz 2 = 3xyz
III. x3 + y 2z + z 2x = 3xyz IV. ( x + y + z )3 = 27 xyz (a) I, IV only (c) I, II and III only
(b) I, II and IV only (d) All of them
What is the perimeter of each of the smaller rectangles? (a) 38
(b) 32
(c) 28
(d) 19
5. The largest nonnegative integer k such that 24k divides 13! is (a) 2
(b) 3
(c) 4
(d) 5
6. In a ∆ ABC, points X and Y are on AB and AC, respectively, such that XY is parallel to BC. Which of the two following equalities always hold? (Here [PQR] denotes the area of ∆PQR).
I. [BCX ] = [BCY ] II. [ ACX ] ⋅ [ ABY ] = [ AXY ] ⋅ [ ABC ]
KVPY Question Paper 2015 Stream : SA
64 (a) Neither I nor II (c) Only II
(b) Only I (d) Both I and II
15. How many ways are there to arrange the letters of
7. Let P be an interior point of a ∆ ABC. Let Q and R be
the word EDUCATION so that all the following three conditions hold?
the reflections of P in AB and AC, respectively. If Q, A, R are collinear, then ∠A equals
— the vowels occur in the same order (EUAIO),
(a) 30°
— no two consonants are next to each other.
(b) 60°
(c) 90°
8. Let ABCD be a square of side length 1, and Γ a circle passing through B and C, and touching AD. The radius of Γ is (a)
3 8
(b)
1 2
(c)
1 2
(d)
5 8
be points in the interiors of the sides AD, BC , AB, CD respectively, such that PQ and RS intersect at right 3 3 , then RS equals angles. If PQ = 4 2 3 2+1 (c) 2
(b)
(a) 15
(b) 24
(c) 72
(d) 120
PHYSICS 16. In an experiment, mass of an object is measured by
9. Let ABCD be a square of side length 1. Let P , Q, R, S
(a)
— the consonants occur in the same order (DCTN),
(d) 120°
3 3 4
applying a known force on it, and then measuring its acceleration. If in the experiment, the measured values of applied force and the measured acceleration are F = 100 . ± 02 . N and a = 100 . ± 001 . m/s 2, respectively. Then, the mass of the object is (b) 10.0 ± 01 . kg (d) 10.0 ± 0.4 kg
(a) 10.0 kg (c) 10.0 ± 0.3 kg
17. A hollow tilted cylindrical vessel of negligible mass
(d) 4 − 2 2
10. In the figure given below, if the areas of the two regions are equal then which of the following is true? 45°
rests on a horizontal plane as shown. The diameter of the base is a and the side of the cylinder makes an angle θ with the horizontal.Water is then slowly poured into the cylinder. The cylinder topples over when the water reaches a certain height h, given by
2y 45°
2y y
y 45°
y 45°
x
h
x
(a) x = y (c) 2x = y
(b) x = 2 y (d) x = 3 y
θ a
11. A man standing on a railway platform noticed that a train took 21 s to cross the platform (this means the time elapsed from the moment the engine enters the platform till the last compartment leaves the platform) which is 88 m long, and that it took 9 s to pass him. Assuming that the train was moving with uniform speed, what is the length of the train in meters? (a) 55 (c) 66
(b) 60 (d) 72
(a) h = 2a tanθ (c) h = a tanθ
18. An object at rest at the origin begins to move in the
+ xdirection with a uniform acceleration of 1 m/s 2 for 4 s and then it continues moving with a uniform velocity of 4 m/s in the same direction.The x t graph for object’s motion will be
n + 1 −3 n
1 be an integer. Which of the following sets of numbers necessarily contains a multiple of 3? (a) n19 − 1, n19 + 1 (c) n38 , n38 + 1
(b) h = a tan 2 θ a (d) h = tanθ 2
(b) n19 , n38 − 1 (d) n38 , n19 − 1
14. The number of distinct primes dividing 12 ! + 13 ! + 14 !
x
4s
t x
(c)
4s
t
(d)
is (a) 5
(b) 6
(c) 7
(d) 8
4s
t
4s
t
65
KVPY Question Paper 2015 Stream : SA 19. If the axis of rotation of the earth were extended into space, then it would pass close to (a) (b) (c) (d)
the moon the sun the pole star the centre of mass of all the planets in the solar system
26. A light bulb of resistance R = 16 Ω is attached in series with an infinite resistor network with identical resistances r as shown below. A 10 V battery drives current in the circuit. What should be the value of r such that the bulb dissipates about 1 W of power. R
20. Methane is a greenhouse gas because (a) it absorbs longer wavelengths of the electromagnetic spectrum while transmitting shorter wavelengths (b) it absorbs shorter wavelengths of the electromagnetic spectrum while transmitting longer wavelengths (c) it absorbs all wavelengths of the electromagnetic spectrum (d) it transmits all wavelengths of the electromagnetic spectrum
21. A parachutist with total weight 75 kg drops vertically onto a sandy ground with a speed of 2 ms −1 and comes to halt over a distance of 0.25 m. The average force from the ground on her is close to
(a) 600 N (c) 1350 N
(b) 1200 N (d) 1950 N
(a) the free electrons in the metal (b) the orbiting electrons of the metal atoms (c) the photons released from the nucleus (d) the nucleus of the metal atoms
23. An optical device is constructed by fixing three identical convex lenses of focal lengths 10 cm each inside a hollow tube at equal spacing of 30 cm each. One end of the device is placed 10 cm away from a point source. How much does the image shift when the device is moved away from the source by another 10 cm? (b) 5 cm
(c) 15 cm
r
10 V
(a) 14.8 Ω
r
(b) 29.6 Ω
r r
r
(c) 7.4 Ω
(d) 3.7 Ω
27. A ball is launched from the top of Mt. Everest which is at elevation of 9000 m. The ball moves in circular orbit around earth. Acceleration due to gravity near the earth’s surface is g. The magnitude of the ball’s acceleration while in orbit is (a) close to g / 2 (c) much greater than g
(b) zero (d) nearly equal to g
28. A planet is orbiting the sun in an elliptical orbit. Let
22. The βparticles of a radioactive metal originate from
(a) 0
r
U denote the potential energy and K denote the kinetic energy of the planet at an arbitrary point on the orbit. Choose the correct statement. (a) K Ualways (c) K =Ualways (d) K =Ufor two positions of the planet in the orbit
29. One mole of ideal gas undergoes a linear process as shown in the figure below. Its temperature expressed as a function of volume V is p0
(d) 45 cm
24. An isosceles glass prism with base angles 40° is clamped over a tray of water in a position such that the base is just dipped in water. A ray of light incident normally on the inclined face suffers total internal reflection at the base. If the refractive index of water is 1.33, then the condition imposed on the refractive index µ of the glass is (a) µ < 2.07 (c) µ < 174 .
(b) µ > 2.07 (d) µ > 174 .
25. A point source of light is moving at a rate of 2 cms −1 towards a thin convex lens of focal length 10 cm along its optical axis. When the source is 15 cm away from the lens, the image is moving at (a) 4 cms−1 towards the lens (b) 8 cms−1 towards the lens (c) 4 cms−1 away from the lens (d) 8 cms−1 away from the lens
(0, 0) V0
(a)
p0V 0 R
(b)
p0V R
(c)
p0V V 1 − R V0
(d)
p0V 0 R
1 −
V V0
2
30. The international space station is maintained in a nearly circular orbit with a mean altitude of 330 km and a maximum of 410 km. An astronaut is floating in the space station’s cabin. The acceleration of astronaut as measured from the earth is (a) zero (b) nearly zero and directed towards the earth (c) nearly g and directed along the line of travel of the station (d) nearly g and directed towards the earth
KVPY Question Paper 2015 Stream : SA
66
42. Among Mg, Cu, Fe, Zn the metal that does not
CHEMISTRY 31. The percentage of nitrogen by mass in ammonium sulphate is closest to (atomic masses of H = 1, N = 14, O = 16, S = 32) (a) 21% (c) 36%
(b) 24% (d) 16%
(b) atomic size (d) electronic configuration
33. Maximum number of electrons that can be accommodated in the subshell with azimuthal quantum number l = 4, is (b) 8
(b) Zn (d) Fe
molecular formula C4 H10 O is
elements are a periodic function of their
(a) 10
(a) Cu (c) Mg
43. The maximum number of isomeric ethers with the
32. Mendeleev’s periodic law states that the properties of (a) reactivity of elements (c) atomic mass
produce hydrogen gas in reaction with hydrochloric acid is
(c) 16
(d) 18
(a) 2
(b) 3
(c) 4
(d) 5
44. The number of electrons required to reduce chromium 3+ completely in Cr2 O2− in acidic medium, is 7 to Cr
(a) 5
(b) 3
(c) 6
(d) 2
45. At constant pressure, the volume of a fixed mass of a gas varies as a function on temperature as shown in the graph
34. The correct order of acidity of the following compounds is
500
OCH3
NO2 V(cm3)
400
COOH 1
COOH 2
300 200
COOH 3
100
(a) 1 > 2 > 3 (b) 1 > 3 > 2 (c) 3 > 1 > 2 (d) 3 > 2 > 1
0
35. Reaction of 2butene with acidic KMnO4 gives (a) CH3CHO (c) CH3CH2OH
(b) HCOOH (d) CH3COOH
100
200
300
T/°C
The volume of the gas at 300°C is larger than that at 0°C by a factor of
36. The gas released when baking soda is mixed with
(a) 3
(b) 4
(c) 1
(d) 2
vinegar is (a) CO
(b) CO2
(c) CH4
(d) O2
37. The element which readily forms an ionic bond has the electronic configuration (a) 1s2 2s2 2 p3 (c) 1s2 2s2 2 p 2
(b) 1s2 2s2 2 p1 (d) 1s2 2s2 2 p 6 3s1
38. The major products of the following reaction, ZnS (s) + O2 ( g ) → ……… are Heat
(a) ZnO and SO2 (c) ZnSO4 and SO2
(b) ZnSO4 and SO3 (d) Zn and SO2
39. If Avogadro’s number is A0, the number of sulphur atoms present in 200 mL of 1N H2SO4 is
A (a) 0 5
A (b) 0 2
A (c) 0 10
(d) A0
40. The functional group present in a molecule having the formula C12 O9 is (a) carboxylic acid (c) aldehyde
(b) anhydride (d) alcohol
41. A sweet smelling compound formed by reacting acetic acid with ethanol in the presence of hydrochloric acid is (a) CH3 COOC2H5 (c) C2H5COOCH3
(b) C2H5 COOH (d) CH3OH
BIOLOGY 46. Excess salt inhibits bacterial growth in pickles by (a) endosmosis (c) oxidation
(b) exosmosis (d) denaturation
47. Restriction endonucleases are enzymes that are used by biotechnologists to (a) cut DNA at specific base sequences (b) join fragments of DNA (c) digest DNA from the 3′ end (d) digest DNA from the 5′ end
48. Enzyme X extracted from the digestive system hydrolyses peptide bonds. Which of the following is probable candidate to be enzyme X ? (a) Amylase (c) Trypsin
(b) Lipase (d) Maltase
49. A person with blood group AB has (a) antigen A and B on RBCs and both antiA and antiB antibodies in plasma (b) antigen A and B on RBCs, but neither antiA nor antiB antibodies in plasma (c) no antigen on RBCs but both antiA and antiB antibodies are present in plasma (d) antigen A on RBCs and antiB antibodies in plasma
67
KVPY Question Paper 2015 Stream : SA 50. Glycolysis is the breakdown of glucose to pyruvic acid. How many molecules of pyruvic acid are formed from one molecule of glucose? (a) 1
(b) 2
(c) 3
(d) 4
51. The process of the transfer of electrons from glucose to molecular oxygen in bacteria and mitochondria is known as (a) TCA cycle (c) fermentation
(b) oxidative phosphorylation (d) glycolysis
52. Which one of the following cell types is a part of innate immunity? (a) Skin epithelial cells (c) Tlymphocytes
(b) Bcells (d) Liver cells
cause impaired blood clotting?
(b) pneumonia (d) filaria
57. Which among grass, goat, tiger and vulture in a food chain, will have the maximum concentration of harmful chemicals in its body due to contamination of pesticides in the soil? (a) Grass since it grows in the contaminated soil (b) Goat since it eats the grass (c) Tiger since it feeds on the goat which feeds on the grass (d) Vulture since it eats the tiger, which in turn eats the goat, which eats the grass
be 500 Da, what is the molecular mass of a doublestranded DNA of 10 base pairs? (a) 500 Da
(b) VitaminC (d) VitaminK
(b) 5 kDa
(c) 10 kDa
(d) 1 kDa
59. Which of the following pairs are both
54. Which one of the following is detrimental to soil fertility? (a) Saprophytic bacteria (c) Nitrobacter
(a) typhoid (c) malaria
58. Considering the average molecular mass of a base to
53. Deficiency of which one of the following vitamins can (a) VitaminB (c) VitaminD
56. Widal test is prescribed to diagnose
(b) Nitrosomonas (d) Pseudomonas
55. In which one of the following phyla is the body
polysaccharides? (a) Cellulose and glycogen (b) Starch and glucose (c) Cellulose and fructose (d) Ribose and sucrose
60. Which one of the following is a modified leaf ?
segmented? (a) Porifera (c) Annelida
(b) Platyhelminthes (d) Echinodermata
(a) Sweet potato (c) Onion
(b) Ginger (d) Carrot
PARTII (2 Marks Questions) MATHEMATICS
64. Given are three cylindrical buckets X , Y , Z whose
61. A triangular corner is cut from a rectangular piece of paper and the resulting pentagon has sides 5, 6, 8, 9, 12 in some order. The ratio of the area of the pentagon to the area of the rectangle is (a)
11 18
(b)
13 18
(c)
15 18
(d)
17 18
62. For a real number x, let [x] denote the largest integer less than or equal to x, and let { x} = x − [x]. The number of solutions x to the equation [x] { x} = 5 with 0 ≤ x ≤ 2015 is (a) 0
(b) 3
(c) 2008
(d) 2009
63. Let ABCD be a trapezium with AD parallel to BC. Assume there is a point M in the interior of the segment BC such that AB = AM and DC = DM. Then, the ratio of the area of the trapezium to the area of ∆ AMD is (a) 2 (b) 3 (c) 4 (d) not determinable from the data
circular bases are of radii 1, 2, 3 units, respectively. Initially water is filled in these buckets upto the same height. Some water is then transferred from Z to X so that they both have the same volume of water. Some water is then transferred between X and Y so that they both have the same volume of water. If hY , hZ denote the heights of water at this stage in h the buckets Y , Z , respectively, then the ratio Y hZ equals (a)
4 9
(b) 1
(c)
9 4
(d)
81 40
65. The average incomes of the people in two villages are P and Q, respectively. Assume that P ≠ Q. A person moves from the first village to the second village. The new average incomes are P′ and Q′, respectively. Which of the following is not possible? (a) P ′ > P and Q ′ > Q (c) P ′ = P and Q ′ = Q
(b) P ′ > P and Q ′ < Q (d) P ′ < P and Q ′ < Q
KVPY Question Paper 2015 Stream : SA
68
70. Stokes’ law states that the viscous drag force F
PHYSICS 66. A girl sees through a circular glass slab (refractive index 1.5) of thickness 20 mm and diameter 60 cm to the bottom of a swimming pool. Refractive index of water is 1.33. The bottom surface of the slab is in contact with the water surface.
experienced by a sphere of radius a, moving with a speed v through a fluid with coefficient of viscosity η, is given by F = 6πηav. If this fluid is flowing through a cylindrical pipe of radius r, length l and a pressure difference of p across its two ends, then the volume of water V which flows through the pipe in time t can be written as a v p = k ηbr c l t where, k is a dimensionless constant. Correct value of a , b and c are
The depth of swimming pool is 6 m. The area of bottom of swimming pool that can be seen through the slab is approximately (a) 100 m 2
(b) 160 m 2
(c) 190 m 2
(d) 220 m 2
67. 1 kg of ice at − 20°C is mixed with 2 kg of water at 90°C. Assuming that there is no loss of energy to the environment, what will be the final temperature of the mixture? (Assume, latent heat of ice = 3344 . kJ/kg, specific heat of water and ice are 4.18 kJ kg −1K −1 and 2.09 kJ kg −1K −1, respectively.) (a) 30°C
(b) 0°C
(c) 80°C
(d) 45°C
68. A rigid body in the shape of a V has two equal arms made of uniform rods. What must the angle between the two rods be so that when the body is suspended from one end, the other arm is horizontal? 1 (a) cos−1 3 1 (c) cos−1 4
(a) a = 1, b = − 1, c = 4 (b) a = − 1, b = 1, c = 4 (c) a = 2, b = − 1, c = 3 (d) a = 1, b = − 2, c = − 4
1 (b) cos−1 2 1 (d) cos−1 6
69. A point object is placed 20 cm left of a convex lens of
focal length f = 5 cm (see in the below figure). The lens is made to oscillate with small amplitude A along the horizontal axis. The image of the object will also oscillate along the axis with
f
CHEMISTRY 71. When 262 g of xenon (atomic mass = 131) reacted completely with 152 g of fluorine (atomic mass = 19), a mixture of XeF2 and XeF6 was produced. The molar ratio XeF2 : XeF6 is (a) 1 : 2 (c) 1 : 1
(b) 1 : 4 (d) 1 : 3
72. Reaction of ethanol with conc. sulphuric acid at 170°C produces a gas which is then treated with bromine in carbon tetrachloride. The major product obtained in this reaction is (a) 1,2dibromoethane (b) ethylene glycol (c) bromoethane (d) ethyl sulphate
73. When 22.4 L of C4 H8 at STP is burnt completely, 89.6 L of CO2 gas at STP and 72 g of water are produced. The volume of the oxygen gas at STP consumed in the reaction is closest to (a) 89.6 L (c) 134.4 L
(b) 112 L (d) 22.4 L
74. The amount of Ag (atomic mass = 108) deposited at the cathode when a current of 0.5 amp is passed through a solution of AgNO3 for 1 h is closest to (a) 2 g (c) 108 g
(b) 5 g (d) 11 g
75. The major product of the reaction is H+/H2O
A
(a) amplitude A / 9 , out of phase with the oscillations of the lens (b) amplitude A / 3, out of phase with the oscillations of the lens (c) amplitude A / 3, in phase with the oscillations of the lens (d) amplitude A / 9, in phase with the oscillations of the lens
Product OH
OH
OH
I
II
(a) I
(b) II
HO III
(c) III
IV
(d) IV
69
KVPY Question Paper 2015 Stream : SA
(c)
enzyme which is a four basepair cutter. What is the frequency with which it will cut the DNA assuming a random distribution of bases in the genome? (a) 1/4
(b) 1/24
(c) 1/256
(d) 1/1296
Time
77. If rice is cooked in a pressure cooker on the Siachen glacier at sea beach and on Deccan plain, which of the following is correct about the time taken for cooking rice?
in animals instead of a monomeric glucose? (a) Energy obtained from glycogen is more than that from the corresponding glucose monomers (b) Glucose present as monomers within the cell exerts more osmotic pressure than a single glycogen molecule, resulting in loss of water from the cells (c) Glucose present as monomers within the cell exerts more osmotic pressure than a single glycogen molecule, resulting in excess water within the cells (d) Glycogen gives more rigidity to the cells
78. A few rabbits are introduced in an uninhabited
80. A line is drawn from the exterior of an animal cell to the centre of the nucleus, crossing through one mitochondrion. What is the minimum number of membrane bilayers that the line will cross?
Population
Population
island with plenty of food. If these rabbits breed in the absence of any disease, natural calamity and predation, which one of the following graphs best represents their population growth?
(b)
Time
79. What is the advantage of storing glucose as glycogen
(a) Gets cooked faster on the Siachen glacier (b) Gets cooked faster at sea beach (c) Gets cooked faster on Deccan plain (d) Gets cooked at the same time at all the three places
(a)
(d)
Population
76. Genomic DNA is digested with Alu I, a restriction
Population
BIOLOGY
(a) 4 (b) 3 (c) 8 (d) 6
Time
Time
Answers PARTI 1
(c)
2
(c)
3
(b)
4
(c)
5
(b)
6
(d)
7
(c)
8
(d)
9
(b)
10
(b)
11
(c)
12
(c)
13
(b)
14
(a)
15
(a)
16
(c)
17
(c)
18
(b)
19
(c)
20
(a)
21
(c)
22
(d)
23
(a)
24
(b)
25
(d)
26
(a)
27
(d)
28
(a)
29
(c)
30
(d)
31
(a)
32
(c)
33
(d)
34
(c)
35
(d)
36
(b)
37
(d)
38
(a)
39
(c)
40
(b)
41
(a)
42
(a)
43
(b)
44
(c)
45
(d)
46
(b)
47
(a)
48
(c)
49
(b)
50
(b)
51
(b)
52
(a)
53
(d)
54
(d)
55
(c)
56
(a)
57
(d)
58
(c)
59
(a)
60
(c)
PARTII 61
(d)
62
(d)
63
(b)
64
(d)
65
(c)
66
(b)
67
(a)
68
(a)
69
(a)
70
(a)
71
(c)
72
(a)
73
(c)
74
(a)
75
(a)
76
(c)
77
(b)
78
(a)
79
(c)
80
(a)
KVPY Question Paper 2015 Stream : SA
70
Solutions 1. (c) We have, f (x) = x2 + ax + 2 and g (x) = x2 + 2x + a Let α be the common root of f (x) = 0 and g (x) = 0. ∴ α 2 + aα + 2 = 0 and α 2 + 2α + a = 0 −α 1 α2 = = ∴ 2 a − 4 a− 2 2− a α2
α = 2 2 −a a −4 ⇒α = and
a 2 − 4 (a + 2) (a − 2) = − (a + 2) = 2− a − (a − 2) α 1 = ⇒α = 1 2− a 2− a
∴
IV. (x + y + z )3 = 27xyz x= y=z Then, (3x)3 = 27x3 Hence option (iv) is also true.
∴(Area of ∆ACX) (Area of ∆ABY ) 1 1 = (AX ) (AC ) sin A × (AY )(AB ) sin A 2 2 1 1 = (AX )(AY ) sin A × (AB )(AC ) sin A 2 2
4. (c) Given, Perimeter of rectangle is 76 units. x
x
y
y
x y
y
x y
y
x
x y
= (Area of ∆AXY ) (Area of ∆ABC) Hence, I and II both are true. y
y x
y
x y
A
…(i) and …(ii) 4x = 3 y On solving Eqs. (i) and (ii), we get x = 6, y = 8 ∴Perimeter of each rectangle = 2 (x + y) = 2 (6 + 8) = 28 units
2. (c) We have,
5. (b) 13! = 2 × 3 × 4 × 5 × 6 × 7
Number of digit of n 2 is 3.
3. (b) We have, x = y = z, x, y, z positive reals. I. x3 + y3 + z3 = 3xyz We know, x3 + y3 + z3 − 3xyz = (x + y + z ) (x2 + y2 + z 2 − xy − yz − zx) 1 = (x + y + z ) [(x − y)2 2 + ( y − z ) 2 + (z − x ) 2 ] When x = y = z Then, (x − y)2 + ( y − z )2 + (z − x)2 = 0 ∴ x3 + y3 + z3 = 3xyz 3 2 II. x + y z + yz 2 = 3xyz Put x = y = z Then, LHS = RHS III. Put x = z = 1and y = 2 Then, it is also true. So, we cannot say only for x = y = z for true
R
Q
Let x and y are sides of each rectangles. ∴Perimetre of rectangle = 6x + 5 y = 76
− (a + 2) = 1 a + 2 = − 1⇒ a = − 3 Now f (x ) + g (x ) = 0 2 ∴x − 3x + 2 + x2 + 2x − 3 = 0 2x2 − x − 1 = 0 1 Q α + β = − b Sum of roots = 2 a n + 2n + 3n + ... + 99n is a perfect square n × 99 × 100 n (1 + 2 + ... + 99), 2 n × 11 × 9 × 2 × 25 = (3)2 × (5)2 × 2 × 11 × n is a perfect square ∴ n must be 22. ∴ n 2 = (22)2 = 484
7. (c) ABC is a triangle. P be interior point of a ∆ ABC, Q and R be the reflections of P in AB and AC respectively.
× 8 × 9 × 10 × 11 × 12 × 13 = 210 × 35 × 52 × 7 × 11 × 13 24k = (23 × 3)k When 13! is divide by 24k 210 × 35 × 52 × 7 × 11 × 13 ∴ 23 k ⋅ 3k = 210 − 3 k ⋅ 35 − k ⋅ 52 × 7 × 11 × 13 ∴ 10 − 3k = integer Then, maximum value of k = 3
6. (d) ABC is a triangle points X and Y on AB and AC respectively. XY is parallel to BC.
θ
φ
θφ P
C
B
QAR are collinear ∴ ∠QAR = 180° Q is reflection of P on AB ∴ ∠QAB = ∠PAB R is reflection of P on AC ∴ ∠RAC = ∠PAC ∠QAR = 180° ∴2 (∠PAB + ∠PAC ) = 180° ∠PAB + ∠PAC = 90° ⇒ ∠BAC = 90°
8. (d) ABCD is a square AB = BC = CD = AD = 1unit C
D r
A N
X
O
1–r
M
Y A
B
r
1 2
C
I. Area of BCX : Area of BCY It is true because same base between same parallels. 1 II. Area of ∆ACX = (AX )(AC ) sin A 2 1 Area of ∆ABY = (AY ) (AB ) sin A 2
B
A circle Γ passing through B and C and touching AD. BC is chord of circle. ∴OM bisects the chord AB 1 1 CM = MB = BC = ∴ 2 2 ⇒ OM = MN − ON = 1 − r
71
KVPY Question Paper 2015 Stream : SA In ∆OMC, OC 2 = OM 2 + CM 2 2
1 r = (1 − r ) + ⇒ 2 1 5 r 2 = 1 − 2r + r 2 + ⇒r = ⇒ 4 8 9. (b) ABCD is square AB = BC = CD = AD = 1 PQ is perpendicular to RS 2
S(5, 1) C(1, 1)
P(0, p) Q(0, q) B(1, 0)
⇒ ⇒
…(i) q− p= r − s (PQ )2 = (1 − 0)2 + (q − p )2
11. (c) Let the length of trains be
15. (a) Given, EDUCATION Vowel occurs in same order _E_U_A_I_O_ There are 6 place for letter DCTN ∴ Total number of arrangement is 6 C4 = 15.
x meter. Time taken by train h cross person = 9 s x ∴ Speed of trains = m/s 9 Time taken by train to cross platform = 21s x x + 88 = ∴ 9 21 [Q length of plateform = 88 m] ⇒ 21x = 9x + 9 × 88 ⇒ 12x = 9 × 88 9 × 88 x= = 66 m ⇒ 12
12. (c) We have, 3
n + 1−3 n
1− 1728 4 12 1 1727 ⇒ n1/3 n1/3 + > 12 432 Put n = 8 only possible least positive integers.
Area of both figures are equal
13. (b) Let n = 3q + r 0≤ r < 3
1 . x (3 y ) 2 3xy 7xy = 2xy + = 2 2
Area of fig. (i) = 2xy +
2y
E
45°
G
D 2x y 45°
45° A
2x (ii)
B
16. (c) As, force F = ma ∆F ∆m ∆a ∆m ∆F ∆a = + ⇒ = − F m a m F a ∆m ± 0.2 ± 0.01 = ⇒ − 10 1 m 0.2 0.01 ∆m So, = + = 0.03 m max 10 1 ⇒
Maximum error in mass occurs when error in force and acceleration are of different signs. So, ∆m = 0.03 × m = 0.03 × 10 ⇒ ∆m= 0.3 kg Hence, mass of object is m = 10 ± 0.3 kg.
17. (c) Cylinder will topple when centre of mass of filled cylinder lies outside the right edge of base. As centre of mass of filled cylinder lies at its midpoint. Cylinder
⇒
10. (b) Given,
F 2x – y 45°
12!(1 + 13 + 13 × 14) 12! (1 + 13 (1 + 14)) 12! × 196 The number of distinct prime of 12! × 196 is 2, 3, 5, 7, 11.
C
∴ n = 3q, 3q + 1, 3q + 2 If n is multiple of 3 i.e. n = 3q Then, n19 is also multiple of 3. When n = 3q + 1and 3q + 2 n38 = (3q + 1)38 = (3q + 1)36 (3q + 1)2 = (36k + 1) (9q2 + 6q + 1) [Q (x + 1)x = nk + 1] 2 = 36k (9q + 6q + 1) + 9q2 + 6q + 1 = 3k + 3λ + 1 ∴ n38 − 1 = 3k + 3λ + 1 − 1 = 3m ∴ n38 − 1 is multiple of 3 Similarly, when n = 3q + 2 n38 − 1 is also multiple of 3.
C h h —— sinθ
θ) /2
A(0, 0) R(r, 0)
Q Slope of PQ × Slope of RS = − 1 q − p 1− 0 ∴ = −1 × 1− 0 s − r
14. (a) We have, 12! + 13! + 14!
sin
D(0, 1)
Area of fig. (ii) Area of ABCG + Area of DEFG = 2xy + (2x − y) y = 2xy + 2xy − y2 = 4xy − y2 7xy 1 = 4xy − y2 ⇒ y2 = xy ⇒ 2y = x ∴ 2 2
(h
2
θ a/2
θ A
B
Now, from above diagram, we have BC h sinθ = ⇒ AC = AC sinθ a So, cosθ = 2 or h = a tanθ h sin θ 2
18. (b) Initially, the velocity is increasing, so the (x  t ) graph must be with increasing slope or parabolic. 1 For first 4s, x = ut + at 2 2 ⇒ x = t 2 / 2 (parabola). After 4 s, particle is moving with a constant velocity, so its graph is a straight line of constant slope after 4 s. After 4 s, velocity is constant. ∴ x = vt = (4 + at )t = 4t (straight line) Hence, best suited option is (b).
KVPY Question Paper 2015 Stream : SA
72 19. (c) Axis of rotation of earth as shown
23. (a) Initially given situation is
below.
10 cm
23.
5°
Polaris (The north star) Axis of rotation
O
Earth’s
30 cm
30 cm
f
2f
f
orbit
10 cm
Perpendicular to plane of orbit
I
20 cm
20 cm
Axis of rotation of earth passes close to polaris, the polar star. O
20. (a) Heat through sun reaches earth
2f
⇒ x2 − rx − x = 0 From sridharacharya formula, we have
20 cm
30 cm
30 cm
10 cm
2f
I
So, distance between object and image in both cases is 90 cm. Hence, there is no shift in image’s position.
24. (b) Base angles of prism is given 40°. So, angle of prism A = 180° − 80° = 100°. 100° 40°
40°
surface, there are two forces acting on her.
µw=1.33
FResistance
Now, before writing Newton’s second law equation, we calculate acceleration of parachutist using v2 − u 2 = 2as 0 − (− 2)2 = 2 (a ) (− 0.25) So, retardation of parachutist is a = 8 ms−2 (directed upwards) Now, using Fnet = ma, we have FR − Fg = ma ⇒ FR − mg = ma or FR = m( g + a ) ⇒ FR = 75(10 + 8) = 1350 N So, resistive force of ground on parachutist is 1350 N.
22. (d) β − particles are emitted from following nuclear reaction: 1 1 0 0 n → 1 p + −1 e + ν A neutron in nucleus is converted into a proton with emission of a β − particle and an antineutrino. This converts emitting nucleus into another nucleus of higher proton number. A 4 0 Z X → z + 1Y + −1 e + ν This decay is characteristics of nuclii for N which, < 1. Z
x= ⇒
f
21. (c) As parachutist lands on earth’s
FGravitation
Adding or removing one of repeating member does not alters the resistance of an infinite network. Let RAB = x, then RAB = RCD . rx x=r+ ⇒ r+ x
When device is moved away from source O, then situation is as shown below.
The southern cross
in form of infrared radiations of higher frequency range approx 1014 Hz. This heat is absorbed by solids on earth’s surface and they reradiate this heat in form of infrared radiations of lower frequency range approx 1010 Hz. These radiations are absorbed by greenhouse gases like methane and does not escapes into space causing warming of earth’s atmosphere.
20 cm
−b ±
x=−
b2 − 4ac 2a
(− r ) ±
r 2 + 4r 2
2 r (1 + 5 ) x= ⇒ 2 Now, power consumed by bulb of resistance R is 1 W, 1 1 ⇒ i= A i 2R = 1 ⇒ i 2 = 16 4 Now, current in circuit is V V ⇒i = i= Rtotal R + RCD 1 10 = ⇒ 4 16 + r (1 + 5 ) 2 r ⇒ 16 + (1 + 5 ) = 40 ⇒ r = 14.8 Ω 2
27. (d) Let orbital radius of ball is r then For TIR, i > θc ⇒ sin 40° > sin θc µ 133 . ≈ 2.07 ⇒ sin 40° > w ⇒ µ g > µg sin 40° µ g > 2.07 25. (d) From mirror formula, we have 1 1 1 …(i) − = v u f or
f = + 10 cm, u = − 15 cm 10 × − 15 = 30 cm v= −15 + 10 Now differentiating Eq. (i) with respect to time, we get dv v2 du = dt u 2 dt
Here, ⇒
dv (+ 30)2 × (+ 2 cm s−1 ) = dt (− 15)2 dv ⇒ = + 8 cm s−1 dt So, image is moving away from lens.
∴
26. (a)
orbital velocity of ball is GM v= r Here, r=R+ h ⇒ r = 6400 km + 9 km or r ≈ 6400 km ⇒ r = R (radius of earth) Now, acceleration of ball in orbit is v2 GM GM = 2 ≈ 2 or a ≈ g a= r r R So, acceleration of ball is nearly equal to g.
28. (a) For a satellite or planet, if total energy is E, then kinetic energy, K = − E and potential energy,U = 2E where, E is negative. So, U> K . 29. (c) Process given is p
R
C r
A r
D
r
r
p0
r
B
V0
V
73
KVPY Question Paper 2015 Stream : SA To find process equation, we use two point form of equation of straight line, y − y1 y − y1 = 2 (x − x1 ) x2 − x1 Here, (x, y,) = (0, p0 ) and (x2 , y2 ) = (V 0 , 0) Process equation is p p = p0 − 0 . V V0 RT As, p= V p RT = p0 − 0 . V ⇒ V V0 T =
⇒
NO2
base
of the compound 3 and
COO–
hence increases the acidity. Thus, the correct order of acidity of the given compounds are
>
p0V V 1 − R V0
COOH 3
30. (d) At height h, acceleration due to gravity is gh =
GM (R + h ) 2
h < < R, GM gh ≈ 2 = g R Direction is towards centre of earth. For
31. (a) Total mass of ammonium sulphate (NH 4 )2 SO4 = 2 × 18 + 32 + 16 × 4 = 36 + 96 = 132 Mass of nitrogen in (NH 4 )2 SO4 = 28 28 % of N by mass in (NH4 )2 SO4 = × 100 132 = 21.2%
32. (c) According to Mendeleev’s periodic law, the physical and chemical properties of the elements are a periodic function of their atomic mass. 33. (d) Maximum number of electrons that can be accommodated in a subshell = 2(2l + 1) When l = 4 Maximum number of electrons = 2(2 × 4 + 1) = 18
34. (c) Electron donating substituents tends to decrease the acidic strength while electron withdrawing substituents tends to increase the acidic strength of substituted benzoic acids relative to benzoic acid. OCH 3 exerts + M effect which destabilises the conjugate base OCH3
of the compound 2 and hence
COO–
decreases the acidity, whereas NO2 exerts −M effect and stabilises the conjugate
OCH3
NO2
40. (b) The structure of C12O9 is as follows O
COOH 1
O COOH 2
KMnO4 /H+ Strong oxidising agent
2 CH3COOH Acetic acid
36. (b) When baking soda is mixed with vinegar aqueous solution of sodium acetate is formed with the evolution of carbon dioxide gas CH3COOH + NaHCO3 Vinegar
[No. of moles = Volume × Molarity] 1 mol of S = A0 atoms 1 A0 atoms. = ∴ 0.1 mole of S = A0 × 10 10
>
35. (d) Acidic potassium permanganate oxides alkenes to ketones or acids depending upon the nature of the alkene. Thus, reaction of 2butene with acidic KMnO4 gives acetic acid. 2 butene
M = 0.5 mol/L ∴No. of moles of H2SO4 = No. of moles of S 0.5 × 200 atom = = 01 . mol 1000
Baking soda
– +
CH3COONa (aq) Sodium acetate
C
C C
C O
38. (a) ∆
2ZnS( s) + 3O 2 ( g ) → 2ZnO + 2SO 2
This process is known as roasting where the sulphide ore is heated in a regular supply of air to give its oxide form at a temperature below the melting point of the metal.
39. (c) Given, Normality of H2SO4 = 1 N Avogadro’s number = A Volume of H2SO4 = 200 mL Normality = Basicity × Molarity For H 2 SO4 , basicity = 2 ∴ 1= 2 × M
O
C O O Mellitic anhydride
Thus, the functional group present in a molecule having C12O9 is an anhydride group.
41. (a) When acetic acid reacts with ethanol in the presence of hydrochloric acid then ethyl acetate (ester) is formed which is a sweet smelling compound. CH3COOH + C2H5OH Acetic acid
Ethanol
CH3COOC2H5 Ethyl acetate (sweet smelling compound)
+ H 2O
Carbon dioxide gas
tendency to form ionic bond as they have low ionisation energy. The general electronic configuration of alkali metal is ns1 . Among the given electronic configuration, 1s2 2s2 2 p 6 3s1 corresponds to the configuration of Na, which is an alkali metal and hence forms ionic bond readily.
O
O
+ H2O (l)+CO2(g)
37. (d) Alkali metals have the highest
Anhydride group
O
C
42. (a) The metals that are present below hydrogen in reactivity series will not produce hydrogen gas in reaction with hydrochloric acid. Among the given metals, Cu is present below H in reactivity series, i.e. it is less reactive than H, will not produce H 2 gas in reaction with HCl acid.
43. (b) Isomers of compound with molecular formula C4 H10 O are as follows CH3 CH2CH2CH2OH CH3 CH(OH)CH2CH3 CH3 — O — CH2CH2CH3 CH3 CH2OCH2CH3 CH3 — O — C H — CH3  CH3 Thus, there are 3 isomeric ethers with molecular formula C4 H10 O. +6
44. (c) C r2O72− + 14H+ → 2Cr3 + + 7H2O As in the above reaction, there are net twelve positive charges on the left side and only six positive charges on right side.
KVPY Question Paper 2015 Stream : SA
74 Therefore, 6 electrons are required to reduce chromium completely in Cr2O72− to Cr3 + in acidic medium Cr2O72− (aq) + 14H+ (aq) + 6e− → 2Cr3 + (aq) + 7H2O
45. (d) 500
V(cm3)
400 300
pyruvate (pyruvic acid) molecules, a total of four ATP molecules and two molecules of NADH.
51. (b) Oxidative phosphorylation is the process in which ATP is formed as a result of the transfer of electrons from NADH or FADH 2 (produced during glycolysis from glucose) to molecular oxygen (O2 ) by a series of electron carriers. It takes place in the mitochondria in eukaryotes and in cytoplasm in prokaryotes.
52. (a) Innate immunity refers to
200 100 0
100
200
300
T/°C
Volume of gas at 0°C V1 = 250 cm3 Volume of gas at 300°C V 2 = 500 cm3 V 2 500 ∴ = =2 V1 250 Thus, the volume of the gas at 300°C is larger than that at 0°C by a factor of 2.
46. (b) Excess salt inhibits growth in pickles by exosmosis. Salt kills and inhibits the growth of microorganisms by drawing water out of the cells of both the microbe and the food through osmosis (or more specifically exosmosis). Due to hypertonic solution outside the bacterial cell, bacteria will die by plasmolysis. 47. (a) Restriction endonuclease is an enzyme that cuts dsDNA into fragments at or near specific recognition sites (palindromic sequence) within the molecule known as restriction sites. These enzymes are found in bacteria and archaea and provide a defence mechanism against invading viruses.
48. (c) In duodenum, trypsin enzyme catalyses the hydrolysis of peptide bonds, breaking down proteins into smaller peptides. Amylase hydrolyses starch into maltose inside the mouth. Lipase breaks down dietary fats into fatty acids and glycerol. Maltase hydrolyses maltose into simple sugar glucose. 49. (b) Person with blood group AB have both A and B antigen in the membrane of his red blood cell but lacks both antibodies (a, b) in his plasma. Due to this reason, blood group AB is called universal recipient.
50. (b) Glycolysis starts with one molecule of glucose and ends with two
nonspecific defence mechanisms that come into play immediately or within hours of an antigen’s appearance in the body. These mechanisms include physical barriers such as skin epithelial cells, chemicals in the blood and immune system cells that attack foreign cells in the body.
53. (d) VitaminK is a cofactor for the enzyme responsible for chemical reactions that maintains blood clotting factors : prothrombin; factor VII, IX, X; and proteins. Thus vitaminK plays a key role in helping the blood clot thereby preventing excessive bleeding.
54. (d) Pseudomonas is denitrifying bacteria. Denitrifying bacteria are microorganisms whose action results in the conversion of nitrates in soil to free atmospheric nitrogen, thus depleting soil fertility and reducing agricultural productivity.
chemicals in its body. Since vulture occupies the top level as it eats the tiger, which eats the goat, which eats the grass in the food chain, it will have the maximum concentration of harmful chemicals in its body.
58. (c) Molecular mass of a base = 500 Da Number of base in a dsDNA = 10 BP or 20 bases Thus, molecular mass of a dsDNA with 20 bases = 20 × 500 = 10 kDa
59. (a) A carbohydrate (e.g. starch, cellulose or glycogen) is a molecule consisting of a number of sugar molecules bonded together by glycosidic linkages and on hydrolysis give its constituent monosaccharides or oligosaccharides. Cellulose is a polymer of β, Dglucose and glycogen of α, Dglucose. Glucose, fructose and ribose are monosaccharides. 60. (c) Onion is a bulb, i.e. it is a modified leaf. A bulb is an underground pyriformspherical structure that possesses a reduced convex or slightly conical discshaped stem and several fleshy scales enclosing a terminal bud. In Onion, the fleshy scales represent leaf bases in the outer part and scale leaves in the central region. 61. (d) We have, A rectangular corner is cut form a rectangular piece of paper. D 3
55. (c) Annelida shows metameric segmentation. It is the repetition of organs and tissues at intervals along the body of an animal, thus dividing the body into a linear series of similar parts or segments (metameres).
56. (a) The widal test is one method used to diagnose enteric fever also known as typhoid fever. Typhoid is caused by Salmonella typhi bacteria. Widal test was based on demonstrating the presence of agglutinin (antibody) in the serum of an infected patient, against the ‘H’ (flagellar) and ‘O’ (somatic) antigens of Salmonella typhi.
57. (d) The increase in concentration of harmful chemical substance like pesticides in the body of living organisms at each trophic level of a food chain is called biological magnification. The organism which occurs at the highest trophic level in the food chain will have the maximum concentration of harmful
4
Q
8
C
5 9
P 6 A
12
B
Area of rectangle = 12 × 9 = 108 sq units Area of pentagon = Area of rectangle − Area of triangle = 108 − 6 = 102 102 17 = ∴ Ratio = 108 18
62. (d) We have,
⇒
∴
[x ] {x } = 5 x ∈[0, 2015] 5 {x } = [x ] {x} ∈ [0,1) 5 5 ∴ Total number of solution is 2009.
75
KVPY Question Paper 2015 Stream : SA 63. (b) Given,
65. (c) Let the number of people in two
ABCD is a trapezium. AD is parallel to BC
villages are x and y respectively. Given, average income of x people = P and average income of y people = Q ∴Total income of people in two villages are Px and Qy respectively.
M is point on BC such that AB = AM and DC = DM
B
A
N
D
P
M
Q
C
In ∆AMD, Area of ∆AMD = Area of ∆AMN + Area of∆DMN Area of ∆AMN = Area of ∆AMP = Area of ∆ABP Area of ∆DMN = Area of ∆DQM = Area of ∆DQC ∴Area of trapezium ABCD = Area of ∆ABM + Area of ∆AMD + Area of ∆MDC = 3 [Area of ∆APM + Area of ∆DMN) = 3 Area of ∆AMD Area of trapezium ABCD ∴ Area of ∆AMD 3 [Area of ∆ADM ] 3 = = = 3:1 Area of ∆ADM 1
64. (d) Let hX , hY and hZ are height of cylindrical bucket of X , Y and Z respectively and rx , ry and rz are radii of bases of cylindrical bucket X , Y and Z respectively. ∴ V X = πrX2 × hX VY = πrY2 × hY V Z = πrZ2 × hZ V X = π hX [Q rx = 1] VY = 4πhY [Q ry = 2] V Z = 9πhZ [Q rz = 3] At initial stage hX = hY = hZ = h ∴ V X = πh, VY = 4πh, V Z = 9πh At second stage some water transfer Z to X, then volume are equal ∴ V X = V Z = 5πh [Q V X + VY = 10πh ] At third stage some water is transferred between x and y. 9 πh V X = VY = [Q V X + VY = 9πh ] 2 Volume of water at third stage 9 VY = 4 πhY = πh 2 and V Z = 9 πhZ = 5 πh 9 πh 4 πhY VY 81 = = 2 = 9 π hZ 5 πh 10 VZ hY 81 = ⇒ 40 hZ
One person moves from first village to second village. Then, number of people in first village = x − 1and second village = y + 1. Average income = P ′ and Q′ ∴Total income = P ′ (x − 1) and Q ′ ( y + 1) Total income in both cases are same ∴ Px + Qy = P ′ (x − 1) + Q ′ ( y + 1) ⇒ Px − P ′ (x − 1) = Q ′ ( y + 1) − Qy ⇒ x(P − P ′ ) + P ′ = y(Q ′ − Q ) + Q ′ ∴ P ′ ≠ P and Q ′ ≠ Q Hence, option (c) is correct. 66. (b) Girl can observe only those light rays which are refracted and leaves the glass slab at angle of 90° or less as shown below. To eye of observer Glass slab
⇒
mw sw ∆T = mi si (0 − (− 20° C)) + mi L + mi sw (T − 0) ⇒ 2 × 418 . × (90 − T ) = 1 × 2.09 × 20 + 1 × 334.4 + 1 × 418 . ×T ⇒ 752.4 − 376.2 = 3 × 418 . ×T ⇒ T = 30°C So, final temperature of mixture is 30°C.
68. (a) Let length of each of rod is l and angle between them is θ.
θ l P Rod 1 A θ C
1
B
Rod 2
D
2
mg
3
r
Pool bottom
x
h=6m
0.6
x
Now, from Snell’s law in layer 1 and 3, we have n1 sin i = n2 sin r 4 1 × sin 90° = × sin r ⇒ 3 3 sin r = 4 Now, from Pythagoras theorem, we have 3 tan r = ⇒ 7 So, from figure, we have x tan r = h 6× 3 = 6.8 m ⇒ x = h tan r = 7 Hence, area of pool visible through glass πd 2 π × (2x + 0.6)2 slab is A = = ≈ 160 m 2 4 4
67. (a) Let final temperature of mixture is T °C. Then, Heat lost by 2 kg water at 90°C to cool down at T°C = Heat gained by 1 kg ice at − 20°C to reach at 0°C + Heat gained by 1 kg ice at 0°C to change its state from ice to water + Water 1 kg formed at 0°C is now absorbs heat to reach temperature of T °C
mg
Let the lower rod is horizontal and upper rod makes θ angle with horizontal. Weights of rods acts vertically downwards from their centres A and B as shown in the above figure. Now, perpendicular distance of weight acting through A from point D is l CD = l cosθ − cosθ 2 l CD = cosθ 2 and perpendicular distance of weight acting through B from point D is l l BD = − l cosθ = (1 − 2 cosθ) 2 2 At equilibrium torque of these two weights about D must balance each other. l l i.e. mg × cosθ = mg × (1 − 2 cosθ) 2 2 3 1 1 ⇒ cosθ = ⇒ cosθ = 2 2 3 −1 1 or θ = cos 3
69. (a) From lens equation, we have 1 1 1 − = v u f Now, differentiating above equation with respect to time, we get dv v2 du dv du = = m2 or dt dt u 2 dt dt
KVPY Question Paper 2015 Stream : SA
76 v = magnification (m) u …(i) ⇒ ∆v = m2 ⋅ ∆u i.e. if object oscillates with an amplitude ∆u, then image also oscillates with amplitude ∆v given by i. Also, magnification, v f …(ii) m= = u f + u
As,
Now, in given question, u = − 20 cm, f = 5 cm f 5 1 So, m= = ⇒ m= − f + u 5 − 20 3
152 = 4 mol 38 2 moles of Xe react completely with 4 moles of F2 to give 1 mol of XeF2 and 1 mol of XeF6. Thus, the molar ratio of XeF2 : XeF6 is 1 : 1. No. of initial moles of F2 =
72. (a) When ethanol reacts with conc. sulphuric acid at 170°C produces ethene gas which is then treated with bromine in carbon tetrachloride to give 1,2dibromoethane as a major product. CH3CH2OH Ethanol
From Eq. (i), we have
CH2
CH2
Ethene
77. (b) The cooking of rice in open
CCl4 (Addition reaction)
CH2—CH2 Br
Br
1, 2 dibromoethane (major)
73. (c) C4 H8 ( g ) + 6O2 ( g ) → 4CO2 ( g ) At STP 22.4 L
89.6 L
+ 4H2O( g ) 72 L
70. (a) By Stokes’ law, F = 6 π η av F We have, η= 6 πav Dimensions of viscosity index η are MLT −2 ⇒ = [ML−1 T −1 ] [ η] = −1 ⋅ L LT Now, given relation of volume flow rate is a p V = k ⋅ ηb ⋅ r c l t Substituting dimensions of physical quantities and equating dimensions on both sides of equation, we have [L3 ] = [ML−2T −2 ]a ⋅ [ML−1 T −1 ]b ⋅ [L]c [T ] ⇒[M0 L3 T −1 ] = [Ma + b L− 2a − b + c T − 2a − b ] Equating dimensions, we have …(i) a+ b= 0 …(ii) − 2a − b + c = 3 …(iii) − 2a − b = − 1 From Eqs. (ii) and (iii), we have c=4 From Eqs. (i) and (iii), we have b= −1 Substituting b in Eq. (i), we have a=1 So, a = 1, b = − 1 and c = 4.
71. (c) 2Xe + 4F2 → XeF2 + XeF6 No. of initial moles of Xe =
170°C (Dehydration of alcohol)
Br2
2
1 ∆v = − × ∆u 3 1 or [given, ∆u = A] ∆v = × A 9 As object is placed between ∞ and 2f distance, so on moving object near to lens, its image moves away from lens. So, oscillations of object and image are out of phase.
Conc. H2SO4
262 = 2 mol 131
So, it is evident that wherever there will be the above sequence in the DNA fragment, Alu I will make blunt cuts over there. Now according to question, if there is random distribution of bases in the genome, the probability of occurrence of the above cut sides will be 1 1 [since Alu I is a 4 base = 4 × 4 × 4 × 4 256 pair cutter] 1 . So, the frequency will be 256
72 = 4 mol 18 1 mole of C4H8 burns completely with 6 moles of O2 to give 4 moles of CO2 and 4 moles of H2O. At STP 1 mole of O2 contains = 22. 4 L 6 moles of O2 contain = 22.4 × 6 = 134.4 L No. of moles of water at STP =
74. (a) Given, current, I = 0.5 A Time, t = 1hr = 3600 s According to Faraday’s IInd law of electrolysis, Atomic mass × I × t W = 96500 108 = × 0.5 × 36000 = 2 g 96500
75. (a)
vessels is favoured at low temperatures and higher altitudes due to the atmospheric pressure. When the rice is cooked in the pressure cooker, then the rice will be cooked faster at the sea beach because the temperature is higher and pressure is lower at sea level than higher altitude. This will allow the water to boil faster inside the pressure cooker and the rice will be cooked faster.
78. (a) In the absence of disease, natural calamities and predation growth of rabbit is exponential. When resources are unlimited, populations exhibit exponential growth, resulting in a Jshaped curve (i.e. option a). 79. (c) Glucose is a monosaccharide and an osmotically active molecule which increases osmotic pressure in cell. So, water enters in cell while glycogen is osmotically inert molecule does not change the osmotic pressure. This is the reason why glucose is not stored in the cell instead glycogen is stored in the animal body.
80. (a) There will be four membrane bilayers that the line will cross 1 = Cell membrane 2 = Mitochondrion 1 = Nucleus
Mitochondrion
The addition of water to alkenes in the presence of an acid form alcohols which occurs through electrophilic addition mechanism and follows Markownikoff rule. Hence, option (a) is correct.
1 2
3
4
Nucleus
76. (c) Alu I has the cut site 5′ AGCT3′ 3′ TCGA5′
Cell membrane
77
KVPY Question Paper 2014 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2014 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS 1. Let r be a root of the equation x + 2x + 6 = 0. The 2
value of (r + 2) (r + 3) (r + 4) (r + 5) is equal to (a) 51
(b) − 51
(c) − 126
(d) 126
2. Let R be the set of all real numbers and let f be a function from R to R such that 1 f (x) + x + f (1 − x) = 1, for all x ∈ R. Then 2 2 f (0) + 3 f (1) is equal to (a) 2
(b) 0
(c) − 2
(d) − 4
coefficient of the highest degree term of k(x) is 1, then sum of the roots of (x − 1) + k(x) is (a) 4
(b) 5
(c) 6
(d) 7
6. In a quadrilateral ABCD, which is not a trapezium, it is known that ∠DAB = ∠ABC = 60°. Moreover, ∠CAB = ∠CBD. Then,
(a) AB = BC + CD (c) AB = BC + AD
(b) AB = AD + CD (d) AB = AC + AD
7. A semicircle of diameter 1 unit sits at the top of a semicircle of diameter 2 units.
3. The sum of all positive integers n for which 13 + 23 + ... + (2n )3 12 + 22 + ... + n 2 (a) 8
(b) 9
is also an integers is (c) 15
1 unit
(d) Infinite
4. Let x and y be two 2digit numbers such that y is obtained by reversing the digits of x. Suppose they also satisfy x2 − y2 = m2 for some positive integer m. The value of x + y + m is (a) 88
(b) 112
(c) 144
(d) 154
5. Let p(x) = x2 − 5x + a and q(x) = x2 − 3x + b, where a and b are positive integers. Suppose HCF ( p(x), q(x)) = x − 1 and k(x) = 1 cm ( p(x), q(x)) If the
2 units
The shaded region inside the smaller semicircle but outside the larger semicircle is called a lune. The area of the lune is (a)
π 3 − 6 4
(b)
3 π − 4 24
(c)
π 3 − 4 12
(d)
3 π − 4 8
KVPY Question Paper 2014 Stream : SA
78 8. The angle bisectors BD and CE of a ∆ ABC are divided by the incentre I in the ratios 3 : 2 and 2 : 1 respectively. Then, the ratio in which I divides the angle bisector through A is (a) 3 : 1 (c) 6 : 5
(b) 11 : 4 (d) 7 : 4
13. The number of 6digit numbers of the form ababab (in base 10) each of which is a product of exactly 6 distinct primes is (a) 8
(b) 10
(c) 13
(d) 15
14. The houses on one side of a road are numbered using
9. Suppose S1 and S2 are two unequal circles, AB and
CD are the direct common tangents to these circles. A transverse common tangent PQ cuts AB in R and CD in S. If AB = 10, then RS is A R Q
B
D
P
consecutive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth house, then (a) 2 ≤ a ≤ 6 (c) 14≤ a ≤ 20
15. Suppose a 2, a3 , a 4 , a5 , a 6 , a7 are integers such that 5 a 2 a3 a 4 a5 a 6 a7 , = + + + + + 7 2! 3! 4! 5! 6! 7! where 0 ≤ a j < j for j = 2, 3, 4, 5, 6, 7. The sum of a 2 + a3 + a 4 + a5 + a 6 + a7 is (a) 8
S
(b) 8 ≤ a ≤ 12 (d) 22≤ a ≤ 30
(b) 9
(c) 10
(d) 11
C
(a) 8
(b) 9
(c) 10
(d) 11
10. On the circle with center O, points A and B are such that OA = AB. A point C is located on the tangent at B to the circle such that A and C are on the opposite sides of the line OB and AB = BC. The line segment AC intersects the circle again at F. Then, the ratio ∠BOF : ∠BOC is equal to A
PHYSICS 16. In the following displacement x versus time t graph, at which among the points P, Q and R is the object’s speed increasing? x P Q (0, 0) R
t
B
(a) R only (c) Q and R only
O F
17. A box when hung from a spring balance shows a C
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 4 : 5
11. In a cinema hall, the charge per person is ` 200. On the first day, only 60% of the seats were filled. The owner decided to reduce the price by 20% and there was an increase of 50% in the number of spectators on the next day. The percentage increase in the revenue on the second day was (a) 50
(b) 40
(c) 30
(d) 20
12. The population of cattle in a farm increases so that the difference between the population in year n + 2 and that in year n is proportional to the population in year n + 1. If the populations in years 2010, 2011 and 2013 were 39, 60 and 123, respectively,then the population in 2012 was (a) 81
(b) 84
(c) 87
(b) P only (d) P, Q and R
(d) 90
reading of 50 kg. If the same box is hung from the same spring balance inside an evacuated chamber, the reading on the scale will be (a) 50 kg because the mass of the box remains unchanged. (b) 50 kg because the effect of the absence of the atmosphere will be identical on the box and the spring balance (c) less than 50 kg because the weight of the column of air on the box will be absent (d) more than 50 kg because the atmospheric buoyancy force will be absent
18. Two positively charged spheres of masses m1 and m2
are suspended from a common point at the ceiling by identical insulating massless strings of length l. Charges on the two spheres are q1 and q2, respectively. At equilibrium, both strings make the same angle θ with the vertical. Then
(a) q1 m1 = q2m2 (c) m1 = m2 sinθ
(b) m1 = m2 (d) q2m1 = q1 m2
79
KVPY Question Paper 2014 Stream : SA 19. A box when dropped from a certain height reaches the ground with a speed v. When it slides from rest from the same height down a rough inclined plane inclined at an angle 45° to the horizontal, it reaches the ground with a speed v / 3. The coefficient of sliding friction between the box and the plane is (Take, acceleration due to gravity is 10 ms −2) (b)
1 9
(c)
2 3
(d)
1 3
20. A thin paper cup filled with water does not catch fire when placed over a flame. This is because (a) the water cuts off oxygen supply to the paper cup (b) water is an excellent conductor of heat (c) the paper cup does not become appreciably hotter than the water it contains (d) paper is a poor conductor of heat
(a) clockwise, while the magnet is above the plane of the ring and counter clockwise, while below the plane of the ring (b) counter clockwise throughout (c) counter clockwise, while the magnet is above the plane of the ring and clockwise, while below the plane of the ring (d) clockwise throughout
27. Two identical bar magnets are held perpendicular to each other with a certain separation, as shown below. The area around the magnets is divided into four zones.
21. Ice is used in a cooler in order to cool its contents.
N
Which of the following will speed up the cooling process? (a) Wrap the ice in a metal foil (b) Drain the water from the cooler periodically (c) Put the ice as a single block (d) Crush the ice
N S
an angle of 60° on the prism, the angle of emergence is 40°. The angle of incidence i for which the light ray will deviate the least is such that (b) 40° < i < 50° (d) i > 60°
23. A concave lens made of material of refractive index 1.6 is immersed in a medium of refractive index 2.0. The two surfaces of the concave lens have the same radius of curvature 0.2 m. The lens will behave as a
Given that there is a neutral point it is located in (a) zone I
25. Two equal charges of magnitude Q each are placed at a distance d apart. Their electrostatic energy is E. A third charge −Q / 2 is brought midway between these two charges. The electrostatic energy of the system is now (b) − E
(c) 0
(d) E
(c)
0
x0
x
Event no.
O
(a) electric field of constant magnitude and varying direction (b) magnetic field of constant magnitude and varying direction (c) electric field of constant magnitude and constant direction (d) electric and magnetic fields of constant magnitudes and constant directions which are parallel to each other
(a) − 2E
(b)
–x0
24. A charged particle initially at rest at O,when released follows a trajectory as shown alongside. Such a trajectory is possible in the presence of
(c) zone III
(d) zone IV
are taken of a particle in a simple harmonic motion between x = − x0 and x = + x0 with origin x = 0 as the mean position. A histogram of the total number of times the particle is recorded about a given position (Event no.) would most closely resemble. (a)
(a) divergent lens of focal length 0.4 m (b) divergent lens of focal length 0.5 m (c) convergent lens of focal length 0.4 m (d) convergent lens of focal length 0.5 m
(b) zone II
28. A large number of random snap shots using a camera
Event no.
22. The angle of a prism is 60°. When light is incident at
(a) i < 40° (c) 50°< i < 60°
S
(d)
–x0
0
x0
Event no.
8 9
through the axis of a copper ring. When viewed from above, the current in the ring will be
x
–x0
0
x0
–x0
0
x0
x
Event no.
(a)
26. A bar magnet falls with its north pole pointing down
29. In 1911, the physicist Ernest Rutherford discovered that atoms have a tiny, dense nucleus by shooting positively charged particles at a very thin gold foil. A key physical property which led Rutherford to use gold was that it was (a) electrically conducting (b) highly malleable (c) shiny (d) nonreactive
30. Consider the following statements: I. All isotopes of an element have the same number of neutrons.
x
KVPY Question Paper 2014 Stream : SA
80 II. Only one isotope of an element can be stable and nonradioactive. III. All elements have isotopes. IV. All isotopes of carbon can form chemical compounds with oxygen16.
Choose the correct option regarding an isotope. (a) Statements III and IV are correct (b) Statements II, III and IV are correct (c) Statements I, II and III are correct (d) Statements I, III and IV are correct
39. At room temperature, the average speed of helium is higher than that of oxygen by a factor of (b) 6 / 2
(a) 2 2
(c) 8
(d) 6
40. Ammonia is not produced in the reaction of (a) NH4Cl with KOH (c) NH4Cl with NaNO2
(b) AlN with H2O (d) NH4Cl with Ca(OH)2
41. The number of isomers which are ethers and having the molecular formula C 4 H10O, is (a) 2
(b) 3
(c) 4
(d) 5
42. The major product of the reaction of 2butene with alkaline KMnO4 solution is
CHEMISTRY 31. The isoelectronic pair is (a) CO, N 2
(b) O2 , NO
(a)
(c) C 2 , HF
hydrazine are, respectively (b) 2 and 6
OH
O
32. The numbers of lone pairs and bond pairs in (a) 2 and 4
(b) O
(d) F2 , HCl (c)
(c) 2 and 5
(d)
(d) 1 and 5
OH
33. The volume of oxygen at STP required to burn 2.4 g of carbon completely is (a) 1.12 L
(b) 8.96 L
OH
43. Among the compounds IIV, the compound having (c) 2.24 L
(d) 4.48 L
the lowest boiling point is
34. The species that exhibits the highest Rf value in a
HO
thin layer chromatogram using a nonpolar solvent on a silica gel plate is
OH
OH
I
II
O (a)
(b)
III N
(c)
(a) I
(d)
(c) III
(d) IV
(i) A r B, ∆G° = 250 kJ mol −1 (ii) D r E, ∆G° = − 100 kJ mol −1
N OH
35. The number of C—C sigma bonds in the compound O
(iii) F r G, ∆G° = − 150 kJ mol −1 (iv) M r N , ∆G° = 150 kJ mol −1 The reaction with the largest equilibrium constant is (a) (i)
is (b) 17
(c) 18
(d) 11
36. If the radius of the hydrogen atom is 53 pm, the radius of the He + ion is closest to (a) 108 pm (c) 27 pm
(b) 81 pm (d) 13 pm (b) NO2
(c) O2
(c) (iii)
(d) (iv)
45. The first ionisation enthalpies for three elements are 1314, 1680 and 2080 kJ mol −1, respectively. The correct sequence of the elements is
(a) O, F and Ne (c) Ne, F and O
(b) F, O and Ne (d) F, Ne and O
46. Individuals of one kind occupying a particular (d) CO2
38. The pH of 0.1 M aqueous solutions of NaCl, CH3COONa and NH4Cl will follow the order (a) NaCl < CH3 COONa < NH4 Cl (b) NH4 Cl < NaCl < CH3 COONa (c) NH4 Cl < CH3 COONa < NaCl (d) NaCl < NH4 Cl < CH3 COONa
(b) (ii)
BIOLOGY
37. The diamagnetic species is (a) NO
(b) II
44. Of the following reactions
+
(a) 16
OH OH IV
geographic area at a given time are called (a) community (c) species
(b) population (d) biome
47. What fraction of the assimilated energy is used in respiration by the herbivores? (a) ~ 10 per cent (c) ~ 30 per cent
(b) ~ 60 per cent (d) ~ 80 per cent
81
KVPY Question Paper 2014 Stream : SA 48. Athletes are often trained at high altitude because (a) training at high altitude increases muscle mass (b) training at high altitude increases the number of red blood cells (c) there is less chance of an injury at high altitude (d) athletes sweat less at high altitude
55. The auditory nerve gets its input from which of the following? (a) The sense cells of the cochlea (b) Vibration of the last ossicle (c) Eustachian tube (d) Vibration of the tympanic membrane
56. Which of the following organelles contain circular
49. In human brain, two cerebral hemispheres are connected by a bundle of fibres which is known as
DNA?
(a) medulla oblongata (c) cerebellum
(a) Peroxisomes and mitochondria (b) Mitochondria and Golgi complex (c) Chloroplasts and lysosomes (d) Mitochondria and chloroplast
(b) cerebrum (d) corpus callosum
50. Which one of the following hormones is produced by the pancreas? (a) Prolactin (c) Luteinising hormone
57. A reflex action does not involve
(b) Glucagon (d) Epinephrine
51. The stalk of a plant leaf is derived from which one of the following types of plant tissue? (a) Sclerenchyma (c) Chlorenchyma
(b) brain (d) muscle fibre
58. Which one of the following options is true in
(b) Parenchyma (d) Collenchyma
photosynthesis?
52. Which of the following muscle types cannot be used voluntarily? (a) Both striated and smooth (b) Both cardiac and striated (c) Both smooth and cardiac (d) Cardiac, striated and smooth
(a) CO2 is oxidised and H 2O is reduced (b) H 2O is oxidised and CO2 is reduced (c) Both CO2 and H 2O are reduced (d) Both CO2 and H 2O are oxidised
59. Human mature Red Blood Cells (RBCs) do not contain (a) iron (c) mitochondria
(b) cytoplasm (d) haemoglobin
60. A person was saved from poisonous snake bite by
53. The pulmonary artery carries
antivenom injection.
(a) deoxygenated blood to the lungs (b) oxygenated blood to the brain (c) oxygenated blood to the lungs (d) deoxygenated blood to the kidney
Which of the following immunities explains this form of protection?
54. Both gout and kidney stone formation is caused by (a) calcium oxalate (c) creatinine
(a) neurons (c) spinal cord
(b) uric acid (d) potassium chloride
(a) Naturally acquired active immunity (b) Artificially acquired active immunity (c) Naturally acquired passive immunity (d) Artificially acquired passive immunity
PARTII (2 Marks Questions) MATHEMATICS
(a) 0
a + b + c = 0, let q = a + b + c and r = a + b + c . Then, 2
2
4
4
4
(a) q2 < 2r always (b) q2 = 2r always (c) q2 > 2r always (d) q2 − 2r can take both positive and negative values 1947
62. The value of
1
∑
2n + 21947
n =0
(a)
487 1945
2
(b)
1946 1947
2
(c)
(c) 2013
(d) 2014
64. In a ∆ ABC with ∠A = 90°, P is a point on BC such
61. Let a , b, c be nonzero real numbers such that 2
(b) 1007
that PA : PB = 3 : 4. If AB = 7 and AC = 5, then BP : PC is (a) 2 : 1
(b) 4 : 3
(c) 4 : 5
(d) 8 : 7
65. The number of all 3digit numbers abc (in base 10) for which (a × b × c) + (a × b) + (b × c) + (c × a ) + a + b + c = 29 is
(a) 6
(b) 10
(c) 14
(d) 18
is equal to
1947 1947
2
(d)
1948 1947
2
63. The number of integers a in the interval [1, 2014] for which the system of equations x + y = a , x2 y2 + = 4 has finitely many solutions is x−1 y−1
PHYSICS 66. A uniform square wooden sheet of side a has its centre of mass located at point O as shown in the figure below on the left. A square portion of side b of this sheet is cut out to produce an Lshaped sheet as shown in the figure on the right.
KVPY Question Paper 2014 Stream : SA
82 b
a
If the refractive index of the material of the prism is µ, then (a) µ > 5 (c) 2 < µ
10 n Sn = (2x + (n − 1)2) 2 Sn = n (x + n − 1)
Now,
⇒
16. (a) In given displacementtime graph, velocity at a particular point is given by the slope of tangent to curve drawn at that point. Speed is the magnitude of velocity, so magnitude of slope gives speed. x
⇒
n=
(a − 11)2 + 680 2
n≥6 Q ⇒ Q
(11 − a ) ±
(a − 11)2 + 680 2
a≤
≥6
800 ≤ 33.33 24
12 ≤ a ≤ 32
a = 12, 14, 16, 18, … When, a = 18, n = 10, then Sn = 170 Q a = 18
15. (b) We have,
θ3
Q
sin
θ
h
θ=45°
As angle of tangent at R(θ3 ) is maximum, so slope’s magnitudem = tan θ3is maximum at R. Hence, speed is increasing at point R.
Net downward acceleration of block is mg sinθ − f a= m
17. (d) In an evacuated chamber, buoyant force of air is absent, so reading of spring balance is more than 50 kg.
where, f = friction force. mg sin θ − µmg cosθ ⇒ a= m
18. (b) In given situation, forces on each of charged sphere are (i) gravitational pull (mg ) kq q (ii) electrostatic repulsion 12 2 r (iii) tension of string (T )
= g (sin θ − µ cosθ) g (1 − µ) = 2 Q sin θ = cosθ = 1 , when θ = 45° 2 Velocity of block when it reaches bottom of inclined plane is
as shown below.
v′ = r2
θ
T1
kq1q2
5 2520a 2 + 840a 3 + 210a 4 + 42a 5 + 7a 6 + a7 = 7! 7
θ
m1
T2
kq1q2 r2
m2
m 1g
T cos θ
T θ
kq1q2 r2
T sin θ
mg
So, T sinθ = ⇒
tanθ =
kq1 q2 r2 kq1 q2
and T cosθ = mg
r 2 ⋅ mg
As angle θ is same for both spheres, we have tan θ1 = tan θ2 kq1 q2 kq q or = 21 2 2 r m1 g r m2 g ⇒
⇒
v′ =
m 2g
If we resolve tension in horizontal and vertical directions, we have following situation in equilibrium.
m1 = m2
2as
where, s = slope length of inclined plane.
=
r
a4 ∈ {1, 2, 3} If a4 = 2, then 2520 + 840 + 210(2) > 3600 Q a4 must be 1 Q 3600 = 2520 + 840 + 210 + 42a5 + 7a6 + a7 ⇒ 30 = 42a5 + 7a6 + a7 a5 ∈ {1, 2, 3, 4} If a5 = 1 30 < 42 + 7a6 + a7 Q a5 = 0 Put a5 = 0, then 30 = 7a6 + a7 Q a6 = 4 and a7 = 2 Q a2 + a3 + a4 + a5 + a6 + a7 = 1+ 1+ 1+ 0 + 4 + 2= 9
mg
R
a2 , a3 , a4 , a5 , a6 , a7 are integers. a a a a 5 a2 a3 = + + 4 + 5 + 6 + 7 7 2! 3! 4! 5! 6! 7! and 0 ≤ a j < j
⇒ 3600 = 2520a2 + 840a3 + 210a4 + 42a5 + 7a6 + a7 0 ≤ aj < j Q a2 = 1 a3 ∈ {1, 2} If a3 = 2, then 2520 + (840) × 2 > 3600 Q a3 must be 1
When block slides down the inclined plane θ = 45°,
θ2
P
(a − 11)2 + 680 2
(11 − a ) ±
2 gh
f
⇒ n 2 + (a − 11)n − 170 = 0 (a − 11) ±
v=
θ1= 0°
170 = n (a − 10 + n − 1)
⇒ n=−
19. (a) When box is dropped from height h, its speed when it reaches the ground is
2 ah / sinθ 2 gh (1 − µ ) × 2
2
=
2 gh (1 − µ ) v = (given) 3 1 So, 2 gh (1 − µ ) = 2 gh 3 1 1− µ = ⇒ 9 1 8 µ = 1− = ⇒ 9 9
20. (c) Thermal resistance of thin layer of paper is quite less, so heat reaches across the paper and water absorbs that heat. Temperature of paper does not rises beyond 100°C and upto its burning temperature. ∴ Paper cup does not catches fire.
21. (d) When ice is crushed, total surface area of ice that comes in contact with surrounding air increases. As a result crushing the ice speed up the cooling process.
87
KVPY Question Paper 2014 Stream : SA 22. (b) Graph of deviation δ versus angle of incidence i is as shown below (for an 3 equilateral glass, µ = prism). 2 δ
kQ 2 kQ 2 / 2 kQ 2 / 2 − − d d /2 d /2 kQ 2 =− =−E d [from Eq. (i)] =
26. (c) Direction of current in ring is given by Lenz’s law, 40°
S
δm=38°
N 40° 45°
50° 60°
i
Clearly, angle of incidence for least deviation lies between 40° < i < 50°.
Current flows anticlockwise in ring when viewed from above
A north pole appears on this face of ring
23. (d) In air, focal length f of concave lens is given by 1 1 1 = (µ − 1) − fair R1 R2 nga = 16 . , R1 = − 0.2 m R2 = + 0.2 m − 2 − 0.6 × 2 1 = (16 . − 1) ∴ = 0.2 0.2 fair 1 ⇒ fair = − m 6 When this lens is dipped in a medium of refractive index nea = 2.0, then fliquid nga − 1 nga − 1 = = fair nge − 1 nga − 1 nea Here, and
1. 6 − 1 0.6 fe f = ⇒ e = . 1 6 fa fa 0.8 − 1 − 1 2 1 0.6 1 fe = × = m = 0.5 m ⇒ 6 0.2 2 Hence, lens acts like a convergent lens of 0.5 m.
When viewed from above current appears clockwise
Current flows anticlockwise in ring when viewed from bottom
S
North pole appears on this face of ring
N
27. (a) Neutral point appears in region in which fields of magnets are in opposite directions.
⇒
N
S
I
II Magnetic fields are in opposite directions in region I
25. (b) Electrostatic energy of two equal charges of magnitude Q placed d distance apart is kq q kQ 2 … (i) E= 1 2= r12 d Q Now, when a third charge − is placed 2 at midpoint of these charges, then electrostatic energy of system is kq q kq2q3 kq q + 1 3 E′ = 1 2 + r12 r23 r13
S III
30. (a) Isotopes have same number of protons. All isotopes of few elements are stable and nonradioactive. Also, all isotopes of few elements are unstable and radioactive. All elements have isotopes. All isotopes of carbon can form compounds with oxygen. So, only statements III and IV are correct. 31. (a, d) Isoelectronic species are those species, which have same number of electrons. The total number of electrons in each pair given in the options are as follows (a) CO, N2 No. of electrons in CO = 6 + 8 = 14 No. of electrons in N 2 = 7 + 7 = 14 (b) O2, NO No. of electrons in O2 = 8 + 8 = 16 No. of electrons in NO = 7 + 8 = 15 (c) C2, HF No. of electrons in C 2 = 6 + 6 = 12 No. of electrons in HF = 1 + 9 = 10 (d) F2, HCl No. of electrons in F2 = 9 + 9 = 18 No. of electrons in HCl = 1 + 17 = 18 Thus, CO and N 2 , F2 and HCl are isoelectronic pairs.
32. (c) The molecular formula of hydrazine is NH 2 NH 2 . From the structure, it is clear that it has 2 lone pairs and 5 bond pairs (4 N — H and 1 N–N).
33. (d) C (s) + O2 ( g ) → CO2 ( g )
N
24. (a) As particle is initially at rest, so to move the charged particle an electric field is required. As path of particle is a curve, so direction of electric field must be changing with distance.
As gold is very malleable, it is possible to produce a foil which is only few atoms thick.
IV
From above figure, we can conclude that magnetic fields cancel each other in region I only.
28. (c) In a simple harmonic motion, oscillating particle passes extreme positions two times, while it crosses mean position once in each half of oscillation. So, graph (c) most closely resembles this situation. 29. (b) In GeigerMarsden experiment, objective is to target αparticles towards an atom. This is possible only when target is a very thin metal foil.
1 mole of carbon reacts completely with 1 mole of oxygen to produce 1 mole of CO2. 12 g of C reacts = 1mole of O2 1 2.4 g of C reacts with = × 2. 4 12 = 0.2 mole of O2 At STP 1 mole of O2 contains = 22. 4 L ∴ 0.2 mole of O2 contains = 22. 4 × 0.2 = 4.48 L 34. (a) The most often used stationary phase gel and alumina are polar material. Consequently, the least polar compound will have the highest Rf value, as they will be least bounded to the stationary phase and moves quickly up the TLC plate. Among the given compound, compound (a) is least polar, so its Rf value will be maximum.
KVPY Question Paper 2014 Stream : SA
88 O
35. (b) 1
6
2
7
8 12
5 3
4
11
9
13
∴ Average speed ∝ 14 16
17
V He = VO 2
15
10
Thus, the number of C — C sigma bonds in the above given compound are 17.
36. (c) According to Bohr’s radius of an atom 52. 9n 2 pm rn = Z where, n = charge on atom Z = atomic number 52.9 × 12 = 26.45 ≈ 27 pm rHe+ = 2 Thus, the radius of He+ ion is closest to 27 pm. 37. (d) Diamagnetic species are those species which have paired electrons in their molecular orbitals. The electronic configurations of molecules given in the options are as follows (i) NO Total number of electrons = 7 + 8 = 15 The electronic configuration of NO will be σ1s2 , σ * 1s2 , σ 2s2 , σ * 2s2 , σ 2 pz2 , π 2 px2 = πzpy2 , π * 2 p1x (ii) NO2 Total number of electrons = 7 + 8 + 8 = 23 The electronic configuration of NO2 will be [18 Ar] σ * 2 pz2 , σ 3s2 , σ * 3s1 (iii) O2 Total number of electrons in O2 = 16 Electronic configuration of O2 will be σ1s2 σ * 1s2 σ 2s2 σ * 2s2 σ 2 pz2 π 2 px2 π 2 py2 π * 2 p1x π * 2 p1y (iv) CO2 Total number of electrons in CO2 = 6 + 8 + 8 = 22 The electronic configuration of CO2 will be [18 Ar] σ * 2 pz2 σ 3s2 Thus, CO2 is a diamagnetic species. 38. (b) NaCl (NaOH + HCl) is a neutral salt, NH4 Cl (HCl + NH4 OH) is an acidic salt while CH3 COONa (CH3 COOH + NaOH) is a basic salt. The value of pH of acidic salt is less than 7, while for neutral salt pH value is equal to 7 and for basic salt pH value is greater than 7. Thus, the increasing order of pH 0.1 M aqueous solutions of NaCl, CH3 COONa and NH4 Cl will be NH4 Cl < NaCl < CH3 COONa. 39. (a) At room temperature 8RT V avg = πM
MO 2 MHe
1 M =
32 = 4
constant. Thus, the reaction with largest equilibrium constant will be F r G, ∆G° = − 150 kJ mol −1
8
45. (a) As O, F and Ne belong to same
V He =2 2 VO 2
⇒
Thus, the average speed of helium is higher than of oxygen by a factor of 2 2.
40. (c) The products formed in each reaction given in the options are as follows (i) NH 4 Cl + KOH → KCl + NH 3 + H 2O (ii) AlN + 3H 2O → Al(OH)3 + NH 3 (iii) NH 4 Cl + NaNO2 → NaCl + N 2 + 2H 2O (iv)NH4Cl + Ca(OH)2 → CaCl 2 + NH3 + H2O Thus, ammonia is not produced in reaction given in option (c). 41. (b) Isomers of compound of molecular formula C4H10O are as follows CH3CH(OH)CH2CH3 CH3CH2CH2CH2OH CH3 — O — CH2 — CH2CH3 CH3 — CH2OCH2CH3 CH3  CH3 — O — C H — CH3
47. (c) The energy assimilated by the
Thus, there are 3 isomers which are ethers having the molecular formula C4 H10 O.
42. (d) The major product of the reaction of 2butene (alkene) with alk. KMnO4 solution is a vicinal glycol, i.e. butane2,3 diol. OH 2 1
3 2butene
4 Alk. KMnO4 (Syn addition) 1
2
period, i.e. 2nd period, the ionisation energies increases on moving from left to right. This increase in ionisation energy is due to decrease in the atomic radii across a period. Thus, the first ionisation enthalpies for O, Fe and Ne are 1314, 1680 and 2080, respectively. So, the correct sequence of the element is O, F and Ne. 46. (b) Population is a group of individuals belonging to same species occupying a particular geographic area in a given time. A community is a group of people living in the same place or having a particular characteristic in common. Biome is a large naturally occurring community of flora and fauna occupying a major habitat. Species is a group of living organisms consisting of similar individuals capable of interbreeding.
3
4
OH Butane2, 3diol
43. (c) The boiling point of a compound depends upon the extent of Hbond present in it. As compound I, II and IV are alcohols, so they can easily form Hbonds while compound III is an ether and cannot form Hbonds. Thus compound III, i.e. has O lowest boiling point. 44. (c) The relation between Gibbs free energy and equilibrium constant can be given as ∆G ° = − 2.303RT log K eq log K eq = − ∆G ° / 2.303RT The reaction having most negative value of ∆G° will have the largest equilibrium
herbivores is used in respiration and a fraction of unassimilated energy is transferred to decomposers (e.g. faecal matter). With increasing trophic levels, the respiration cost also increases sharply. On an average, producers consume about 20% of their gross productivity in respiration. The herbivores consume about 30% of assimilated energy in respiration. In carnivores, the proportion of assimilated energy consumed in respiration rises to about 60%.
48. (b) Athletes are often trained at high altitude because the air is ‘thinner’ at high altitudes, means there are fewer oxygen molecules per volume of air. Every breath taken at high altitude delivers less of what working muscles require. To compensate for the decrease in oxygen there occurs more production of red blood cells to aid in oxygen delivery to the muscles. 49. (d) Corpus callosum is nervous band which attaches both cerebral hemispheres of mammals. It is a thick band of nerve fibres that divides the cerebral cortex lobes into left and right hemispheres. It connects the left and right sides of the brain allowing for communication between both hemispheres.
89
KVPY Question Paper 2014 Stream : SA 50. (b) Glucagon is secreted from alpha cells of pancreas. Prolactin is secreted by the anterior pituitary. Luteinising hormone is secreted by the gonadotropic cells in the anterior pituitary. Epinephrine (or adrenaline) is secreted by the medulla of the adrenal gland.
51. (d) The stalk of plant leaf (petiole) is derived from collenchyma. Collenchyma cells are elongated cells with irregular thick cell walls that provide structural support, particularly in growing shoots and leaves. Their thick cell walls are composed of the compounds cellulose and pectin.
52. (c) Both smooth muscle (unstriated muscle) and cardiac muscle are functionally involuntary. Muscles that are under our conscious control are called voluntary muscles, while muscles that are not under our conscious control are called involuntary muscles. Striated muscles are voluntary muscles.
53. (a) Pulmonary artery arises from left ventricle and carries deoxygenated blood to the lungs. The pulmonary artery begins in the heart at the base of the right ventricle.
54. (b) Gout is caused by the deposition of uric acid in joints. Composition of kidney stone is calcium oxalate, calcium phosphate, uric acid, xanthine and indigo calculi. Researches have shown that kidney stones are a complication of gout because extra uric acid can collect in the urinary tract and crystallise into stones.
55. (a) Cochlea is the main hearing organ. It is composed of sensory cells called hair cells, which convert vibrations into neural messages. These messages are then passed to the auditory nerve and carried up to the brain.
56. (d) Mitochondria and chloroplast contain circular DNA. Even though both organelles are found in eukaryotic cell, both mitochondria and chloroplast have characteristics often found in prokaryotic cells. These prokaryotic cell’s characteristics include enclosed double membrane, circular DNA and bacteria like ribosomes.
Its pathway is as discussed below Receptor → Sensory neuron → (Skin) Integration centre
1947
∑
n=0
(Spinal cord)
 ↓ Effector ← Motor neuron
Light reaction → Photolysis of water (H 2O is oxidised) 2H 2O → O2 + 4 [H] requires light reaction
=
59. (c) A mature RBC lacks nucleus, mitochondria and endoplasmic reticulum. In humans, mature RBCs are flexible and oval biconcave disks. They lack a cell nucleus and most organelles, in order to accommodate maximum space for haemoglobin.
60. (d) Antivenom injection provides
⇒ (a 2 + b2 + c2 )2 = a 4 + b4 + c4 + 2 [(ab + bc + ca )2 − 2abc (a + b + c)] 2 ⇒ q = r + 2[(ab + bc + ca )2 − 2(abc)(0)] ⇒ q2 = r + 2 [ab + bc + ca ]2 ⇒ q2 = r + 2
⇒ ⇒
=
+ 1
1 1947 2 2
1 1947 2 2
1947
Q
∑ f (x) = f (0) + f (1) + f (2) + f (3) + .... + f (1947) = (f (0) + f (1947) + (f (1) + f (1946)) + ... ... + (f (973) + f (974)) 1 = 974 × 1947
1947
∑
n=0
2
f (n ) =
2 2 2 × 487 2
1945 ×2 2
=
487 21945
63. (d) Given, x + y = a ⇒ a∈[1, 2014]
∴
+ 2 (a 2b2 + b2c2 + c2a 2 )
1 q − q2 = r 2 q2 = 2r
+1
1947 1947 2 2 2 2
Similarly, f (1) + f (1946) =
and
⇒ a 2 + b2 + c2 = q ⇒ a 4 + b4 + c4 = r ⇒ (a 2 + b2 + c2 )2 = a 4 + b4 + c4
2
1947
21947 + 2 2 1 + 1947 1947 2 2 2 2 + 1 2
61. (b) Given, a + b + c = 0 ⇒ a, b, c ∈ R
2q2 4
1947 2
2
1
n=0
artificial acquired passive immunity. Passive immunity is the transfer of active humoral immunity of readymade antibodies. It can occur naturally, when maternal antibodies are transferred to foetus through placenta and it can be induced artificially when high level of antibodies specific to a pathogen or toxin are transferred to nonimmune persons through blood products that contain antibodies, such as antivenom injections are given.
q2 = r +
1947 2
1+ 2
4 [H] + CO2 → (CH 2O) + H 2O (Reduction)
⇒
1 n
1+ 2 1
=
Dark reaction → CO2 is reduced for sugar formation
⇒
21947
2 + 21947 1 f (0) + f (1947) = + 1947
(muscle)
(a + b + c)2 − (a 2 + b2 + c2 ) 2 2 0 − q 2 q = r + 2 2
1 2n +
f (n ) =
Let
58. (b) In photosynthesis,
57. (b) Reflex actions do not involve the brain in the decision making process. Reflex action is a rapid, spontaneous and involuntary activity that is produced in response to a stimulus. It is controlled by spinal cord.
62. (a) We have,
x2 y2 + =4 x−1 y−1 x2 y2 + =4 x −1 y −1
⇒ x2 y − x2 + xy2 − y2 = 4(x − 1)( y − 1) ⇒ x2 y + xy2 = 4(xy − (x + y) + 1) + x2 + y2 ⇒ xy (x + y) = 4(xy − a + 1) + (x + y)2 − 2xy ⇒
xy(a ) = 4xy − 4a + 4 + a 2 − 2xy
⇒ xya − 2xy = a 2 − 4a + 4 ⇒ xy (a − 2) = (a − 2)2 ⇒
(a − 2)2 − xy (a − 2) = 0
⇒
(a − 2) (a − 2 − xy) = 0
⇒
a = 2 or xy = a − 2
or
x (a − x ) = a − 2
⇒
x2 − ax + a − 2 = 0
[Q y = a − x]
KVPY Question Paper 2014 Stream : SA
90
⇒ (a × b)(c + 1) + b (c + 1) + a (c + 1) + (c + 1) = 30 ⇒ (c + 1) (a × b + b + a + 1) = 30 ⇒ (c + 1) (b(a + 1) + 1(a + 1)) = 30 ⇒ (a + 1)) (b + 1) (c + 1) = 30 1 ≤ a ≤ 9, 0≤ b, c ≤ 9 Q Total number of solution = 18
Since, x ∈ R D≥0
Q
a 2 − 4 (a − 2) ≥ 0
Q
a − 4a + 8 ≥ 0, ∀ a ∈ R 2
a ∈ [1, 2014]
Q
64. (a) Given, ABC is right angled triangle. B
66. (b) Centre of mass of square wooden plate with respect to chosen axis is at centre of plate.
4x
O
x
P
a 2
√7
Rearranging, we get 2
a a ⇒ − − 1 = 0 b b a 1± 1+ 4 = b 2 5+1 a ⇒ = 2 b Note Choice of a different origin gives a a different value of . b ⇒
67. (d) For a soap bubble floating in air, Gravitational force = Buoyant force ⇒ g (mass of helium + mass of soap film) = Weight of air displaced by bubble …(i)
a/2
3x
√5
C
t
∠A = 90° AC = 5 AB = 7 PA 3 = PB 4 In ∆ABC BC 2 = AB 2 + AC 2 = 5 + 7 = 12 BC = 12 = 2 3 In ∆ABP, AB 2 + PB 2 − AP 2 cosB = 2AB ⋅ PB 7 + 16x2 − 9x2 7 ⇒ = 2 3 2 ⋅ 7 × 4x 7 AB Q cosB = BC = 2 3 ⇒ ⇒ ⇒ ⇒ Q
y
A
28x = 7 + 7x2 ⇒ 28x = 7 3 (x2 + 1) 3 3 x2 − 4x + 3 = 0 3x − 3x − x + 3 = 0 ( 3x − 1) (x − 3 ) = 0 1 ,x≠ 3 x = 3, 3 2
PB = 4 / 3 4 2 PC = BC − BP = 2 3 − = 3 3 BP 4 / 3 = = 2:1 Q PC 2 / 3
r
Coordinates of centre of mass are a a x1 = , y1 = 2 2 Mass of square plate, m1 = ka 2 where, k = mass per unit area. Coordinates of centre of mass of removed portion and its mass with respect to axes chosen are O
x b/2 b/2
y
b b , y2 = 2 2 and mass of removed portion, m2 = kb2 Now, centre of mass of remaining Lshaped portion is given at point P whose coordinates are (b, b) with respect to chosen axes. x2 =
68. (c) As heat lost by aluminium piece = heat gained by water
x
Q
{ms (Ti − Tf )}aluminium = {ms(Tf − Ti )}water P(b,b)
Substituting given values, we get ⇒
50 × 10−3 × 900 × (300 − 160)
⇒
6300 = 4200 (T − 30)
= 1 × 4200 × (T − 30)
65. (d) abc is threedigits number abc = 100a + 10b + c 100 ≤ abc < 999, a ∈ {1, 2, 3, ..., 9} b, c∈{0, 1, 2, 3, ..., 9} Now, (a × b × c) + (a × b) + (b × c) + (c × a ) + a + b + c = 29
Let r = inner radius of soap bubble and t = thickness of film. Then, from Eq. (i), we have 4 3 ⇒ πr × ρHe + 4 πr 2 × t × ρsoap 3 4 = πr3 × ρair 3 Substituting values in above equation, we get 4 . + 4 π × (10−2 )2 ⇒ × π × (10−2 )3 × 018 3 × t × 1000 4 = × π × (10−2 )3 × 1. 23 3 Rearranging, we get 4 . ) ⇒ 4 π (10−2 ) ⋅ t ⋅ 1000 = π (10−6 ) (108 3 ⇒ (10 5 ) t = 0.35 or t = 3.5 × 10−6 m = 3.50 µm
y
Now, using XCM =
m1 x1 − m2x2 , we have m1 − m2
a b ka 2 − kb2 2 2 a3 − b3 b= ⇒b = 2 2 2(a 2 − b2 ) ka − kb
⇒
T = 30 + 1.5
or
T = 31.5°C
So, temperature of water after taking out aluminium piece is 31.5°C.
91
KVPY Question Paper 2014 Stream : SA 5 i4 = i3 3
69. (a) From the geometry of given figure,
8 (3i1 ) = 8 i1 3 Similarly, across section bb, =
45°+ r 45°+ r
r 45°– r 45°– r
45°
i7
We have, for total internal reflections …(i) 45° + r > θc and …(ii) 45° − r > θc or 90° > 2θc ⇒ sin 45° > sin θc 1 1 …(iii) > or µ > 2 ⇒ 2 µ
µ2 − 1 µ
−
i1
d X
b
c
d
Now, we consider section dd, i1 R
i3 d i2 R
R d
Equating potential across dd, we get i2R = i1 (2R ) or i2 = 2i1 Hence, current i is i3 = i1 + i2 = i1 + 2i1 = 3i1 Now, we consider section cc, i5
i3 d
c i4 R
2R × R 2R = 3 2R + R
R c
b 13 R 21
a
1 mole of NH3 = 17 g ∴ 4 × 10−2 moles of NH3 = 17 × 4 × 10−2 ∴ WNH 3 = 17 × 4 × 10−2 g 14 × 17 × 4 × 10−2 ⇒ WN = 17 = 0.56 g 0.56 % of N = × 100 = 28% 2
b
13 + Ri8 = i7 21
1 R
73. (d) Let the equivalent weight of
34 Ri8 = i7 R 21 34 i8 = i7 × ⇒ 21 Hence, current i is i = i7 + i8 34 55 i7 = i7 + i7 = 21 21 55 = × 21i1 21 = 55i1 = 55 × 10−3 A (Q i1 = 1 mA, given) Total resistance across PQ is 34 Req = kΩ 55 34 × 1000 Ω = 55 So, potential drop across, PQ 34 = i Req = 55 × 10−3 × × 103 V 55 = 34 V
metal M = x (eq )Ca (eq )H 2 , released = (eq )H 2 , released (eq )M
⇒
∴ ⇒
. 1125 2 ( eq vol )H 2 20 = . 2 185 (eq vol. )H 2 x 1.125 x = 20 1.85 1.125 x= × 20 1.85 = 12.16 ≈ 12
74. (a) When coke is heated with lime (CaO), then CaC2(X ) is formed which then reacts with water to form acetylene (Y ) as a major product. This acetylene on passing over red hot iron at 873 K produces benzene (Z). CaO + C
71. (b) 18 g of H2O = 1 mole
∆
(Coke)
1 × 3.6 = 0.2 mole 18 2H2 ( g ) + O2 ( g ) → 2H2O(l)
CaC2 X
Calcium carbide
3.6 of H2O =
d
Equating potentials, we get 2 i4 R = i3 R + R 3
[basicity of H2SO4 = 2] Eq. of H2SO4 = Eq. of NH3 = No. of moles of ammonia = 4 × 10−2
R
R
be i.
Q a
i7 i8
70. (d) Let current through resistor X is c
Eq. of H2SO4 = 2 × 2 × 10 × 10−3
c
b
⇒
72. (a) 2NH3 + H2SO4 → (NH4 )2 SO4
5 R 8
13 13 Ri6 = R i5 or i6 = i5 8 8
1 1 > ⇒ µ2 − 1> 2 µ µ
b
c
R
i a
…(iv) ⇒ µ> 5 Common solution of Eqs. (iii) and (iv) is µ > 5.
P a
b i5 i6 R
So, current, 13 21 21 i5 = (8i1 ) = 21 i1 i7 = i5 + i5 = 8 8 8 Now, for section aa, we have
From Eq. (ii), we have 45 − θc > r sin(45 − θc ) > sin r 1 1 sin 45° cosθc − sin θc > ⇒ 2 2 µ ⇒
5 8 i3 + i3 = i3 3 3
So, current, i5 = i4 + i3 =
45°
∴ Resulting moles of gases in mixture = x − 0.2 + 10 − x − 01 . = 9.7 At constant temperature and volume, p1 p = 2 n1 n2 p 1 = 2 ⇒ 10 9.7 ⇒ p2 = 0.97 atm
Initially
x mol
After reaction x − 0.2
(10 − x ) mol 0.2 10 − x − 2
0 0.2 mol
H 2O Fe (red hot) 373 K (Trimerisation)
Z Benzene
CH Y
CH + Ca(OH)2
Acetylene
KVPY Question Paper 2014 Stream : SA
92
The ETC occurs in the plasma membrane of prokaryotes and the inner mitochondrial membrane of eukaryotes.
75. (a) Br
(i) Alc. KOH Ph (ii) NaNH2
Br
77. (c) Colourblindness is Xlinked
(Dehydrohalogenation)
H
C
C
recessive disorder. Ph
(X) HgSO4/dil. H2SO4, ∆ (Acidic hydration of alkyne) Tautomerism
CH3
C
H 2C
C
Ph
XXC Heterozygous woman
XXC
Conc. HNO3/H2SO4 nitration
NO2
O
(misomer)
(Y)
76. (c) Respiratory reactions occur in the mitochondria and cytoplasm. Respiratory reactions include glycolysis, Kreb’s cycle and Electron Transport Chain (ETC). The glycolysis always occurs in the cytoplasm of all living cells. The Kreb’s cycle occurs in the cytoplasm of all prokaryotes and in the mitochondrial matrix in eukaryotes.
XCXC
XY
XCY
79. (a) Phenylketonuria is an autosomal recessive disorder with mutation in gene for enzyme Phenylalanine Hydroxylase (PAH), making it nonfunctional. PAH
Phenylalanine → Tyrosine
Ratio
OH
O
XCY Colourblind man
×
On the other hand, the second beaker contains 10% sucrose solution which is more concentrated than the semipermeable bag’s sucrose concentration. Hence, water will move out of the bag to the beaker and make the bag flaccid (exosmosis). Thus, the answer (b) is correct.
X
Genotype F1 ratio Carrier daughter
XXC
1
Colourblind daughter
XC XC
1
Normal son
XY
1
Colourblind son
XC Y
1
78. (b) Osmosis is the net movement of solvent molecules into a region of higher solute concentration through a semipermeable membrane. In the given question, one of the 2 sucrose containing bags (semipermeable) is placed in a water containing beaker. Clearly, the concentration in the bag is more than the beaker and as a result, water will move through the semipermeable membrane (endosmosis) and make the bag turgid.
Such person cannot metabolise the above reaction leading to accumulation of phenylalanine. So, are given food low in phenylalanine and supplemented with tyrosine.
80. (a) Ganga is more polluted than Kaveri → lower DO [Dissolved Oxygen] indicates polluted water 2%
Ganga water (P)
10%
Kaveri water (Q)
DO
Dissolved oxygen refers to the level of free oxygen present in water levels that are too high or too low can harm aquatic life and affect water quality.
93
KVPY Question Paper 2013 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2013 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS
5. Let x, y, z be nonzero real numbers such that x y z y z x x3 y3 z3 + + = 7 and + + = 9, then 3 + 3 + 3 − 3 y z x x y z y z x is equal to
1. Let x, y, z be three nonnegative integers such that x + y + z = 10. The maximum possible value of xyz + xy + yz + zx is (a) 52
(b) 64
(c) 69
(a) 152
(d) 73
2. If a , b are natural numbers such that 2013 + a = b , 2
then the minimum possible value of ab is (a) 671 (c) 658
(b) 668 (d) 645
3. The number of values of b for which there is an
isosceles triangle with sides of lengths b + 5, 3b − 2 and 6 − b is
(a) 0
(b) 1
(c) 2
(d) 3
4. Let a , b be nonzero real numbers. Which of the following statements about the quadratic equation ax2 + (a + b)x + b = 0 is necessarily true? I. It has at least one negative root. II. It has at least one positive root. III. Both its roots are real. (a) I and II only (c) II and III only
(b) I and III only (d) All of them
2
(b) 153
(c) 154
(d) 155
6. In a ∆ ABC with∠A < ∠B < ∠C, points D, E , F are on the interior of segments BC , CA, AB respectively. Which of the following triangles cannot be similar to ∆ABC? (a) ∆ ABD
(b) ∆ BCE
(c) ∆CAF
(d) ∆ DEF
7. Tangents to a circle at points P and Q on the circle intersect at a point R. If PQ = 6 and PR = 5, then the radius of the circle is 13 3 15 (c) 4
(a)
(b) 4 (d)
16 5
8. In an acute angled ∆ ABC, the altitudes from A, B, C when extended intersect the circumcircle again at points A1 , B1 , C1 respectively. If ∠ABC = 45°, then ∠A1B1C1 equals (a) 45°
(b) 60°
(c) 90°
(d) 135°
94
KVPY Question Paper 2013 Stream : SA
9. In a rectangle ABCD, points X and Y are the midpoints of AD and DC, respectively. Lines BX and CD when extended intersect at E, lines BY and AD when extended intersect at F. If the area of ABCD is 60, then the area of BEF is (a) 60
(b) 80
(c) 90
(d) 120
10. In the figure given below, ABCDEF is a regular hexagon of side length 1, AFPS and ABQR are squares. Then, the ratio ar ( APQ) /ar (SRP ) equals A
B
PHYSICS 16. A man inside a freely falling box throws a heavy ball towards a side wall. The ball keeps on bouncing between the opposite walls of the box. We neglect air resistance and friction. Which of the following figures depicts the motion of the centre of mass of the entire system (man, the ball and the box)? (a)
(c)
(b)
(d)
S F
R
Q
17. A ball is thrown horizontally from a height with a
C
certain initial velocity at time t = 0. The ball bounces repeatedly from the ground with the coefficient of restitution less than 1 as shown below.
P E
(a)
2+1 2
(b) 2
D
3 3 (c) 4
(d) 2
11. A person X is running around a circular track completing one round every 40 s. Another person Y running in the opposite direction meets X every 15 s. The time, expressed in seconds, taken byY to complete one round is (a) 12.5
(b) 24
(c) 25
(d) 55
Neglecting air resistance and taking the upward direction as positive, which figure qualitatively depicts the vertical component of the ball’s velocity vy as a function of time t? vy
12. The least positive integer n for which n + 1 − n − 1 < 02 . is (a) 24 (c) 26
(b) 25 (d) 27
vy
(a)
t
(b)
t
13. How many natural numbers n are there such that n ! + 10 is a perfect square?
(a) 1 (c) 4
(b) 2 (d) infinitely many
vy
vy
t
(c)
t
(d)
14. Ten points lie in a plane so that no three of them are collinear. The number of lines passing through exactly two of these points and dividing the plane into two regions each containing four of the remaining points is (a) 1 (b) 5 (c) 10 (d) dependent on the configuration of points
18. A tall tank filled with water has an irregular shape as shown. The wall CD makes an angle of 45° with the horizontal, the wall AB is normal to the base BC. The lengths AB and CD are much smaller than the height h of water (figure not to scale).
15. In a city, the total income of all people with salary below ` 10000 per annum is less than the total income of all people with salary above ` 10000 per annum. If the salaries of people in the first group increases by 5% and the salaries of people in the second group decreases by 5%, then the average income of all people (a) increases (b) decreases (c) remains the same (d) cannot be determined from the data
h p3
A
B
p1
D
p2
C
95
KVPY Question Paper 2013 Stream : SA Let p1, p2 and p3 be the pressures exerted by the water on the wall AB, base BC and the wall CD respectively. Density of water is ρ and g is acceleration due to gravity. Then, approximately (b) p1 = 0, p3 =
(a) p1 = p2 = p3 (c) p1 = p3 =
1 p2 2
1 p2 2
(d) p1 = p3 = 0, p2 = hρg
23. In a car, a rear view mirror having a radius of curvature 1.50 m forms a virtual image of a bus located 10.0 m from the mirror. The factor by which the mirror magnifies the size of the bus is close to (a) 0.06
(b) 0.07
(c) 0.08
(d) 0.09
24. Consider the following circuit shown below. I I′
19. The accompanying graph of position x versus time t represents the motion of a particle. If p and q are both positive constants, the expression that best describes the acceleration a of the particle is E
x
All the resistors are identical. The ratio of I / I ′ is (a) 8 (b) 6 (c) 5 (d) 4
25. The figure shows a bar magnet and a metallic coil.
t
(a) a = − p − qt (c) a = p + qt
(b) a = − p + qt (d) a = p − qt
Consider four situations: (I) Moving the magnet away from the coil. (II) Moving the coil towards the magnet. (III) Rotating the coil about the vertical diameter. (IV) Rotating the coil about its axis.
20. Two stones of masses m1 and m2 (such that m1 > m2)
are dropped ∆t time apart from the same height towards the ground. At a later time t, the difference in their speed is ∆v and their mutual separation is ∆s. While both stones are in flight
(a) ∆v decreases with time and ∆s increases with time (b) Both ∆v and ∆s increase with time (c) ∆v remains constant with time and ∆s decreases with time (d) ∆v remains constant with time and ∆s increases with time
An emf in the coil will be generated for the following situations. (a) I and II only (c) I, II, and III only
26. A current of 0.1 A flows through a 25 Ω resistor represented by the circuit diagram. The current in 80 Ω resistor is 80Ω
21. The refractive index of a prism is measured using
25Ω
three lines of a mercury vapour lamp. If µ1 , µ 2 and µ3 are the measured refractive indices for these green, blue and yellow lines respectively, then
(a) µ 2 > µ 3 > µ 1 (c) µ 3 > µ 2 > µ 1
vertical convex lens of focal length 20 cm and is then reflected by a tilted plane mirror, so that it converges to a point I. The distance PI is 10 cm. I
P
M is a point at which the axis of the lens intersects the mirror. The distance PM is 10 cm. The angle which the mirror makes with the horizontal is (a) 15°
(b) 30°
(c) 45°
(d) 60°
20Ω 20Ω
(a) 0.1 A
(b) 0.2 A
60Ω
(c) 0.3 A
(d) 0.4 A
27. Solar energy is incident normally on the earth’s
surface at the rate of about 1.4 kW m −2. The distance between the earth and the sun is 15 . × 1011 m. Energy E and mass m are related by Einstein equation E = mc2, where c = 3 × 108 ms−1 is the speed of light in free space. The decrease in the mass of the sun is
(a) 109 kg s−1 (c) 1026 kg s−1
M
0.1 A
V
(b) µ 2 > µ 1 > µ 3 (d) µ 1 > µ 2 > µ 3
22. A horizontal parallel beam of light passes through a
(b) I, II and IV only (d) I, II, III, and IV
(b) 1030 kg s−1 (d) 1011 kg s−1
28. If the current through a resistor in a circuit increases by 3%, then the power dissipated by the resistor (a) increases approximately by 3% (b) increases approximately by 6% (c) increases approximately by 9% (d) decreases approximately by 3%
96
KVPY Question Paper 2013 Stream : SA
The gas is compressed isothermally to the volume V / 3. Now, the cylinder valve is opened and the gas is allowed to leak keeping temperature same. What percentage of the number of molecules should escape to bring the pressure in the cylinder back to its original value? (a) 66%
(b) 33%
(c) 0.33%
36. Which of the following molecules has no dipole moment? (a) CH 3 Cl
(b) CHCl 3
(c) CH 2 Cl 2
(d) CCl 4
37. The decay profiles of three radioactive species A, B and C are given below :
Concentration
29. An ideal gas filled in a cylinder occupies volume V .
(d) 0.66%
30. An electron enters a chamber in which a uniform magnetic field is present as shown below.
A B
C
Time
These profiles imply that the decay constants kA , kB and kC follow the order (b) kA > kC > kB (d) kC > kB > kA
(a) kA > kB > kC (c) kB > kA > kC
e–
38. A specific volume of H2 requires 24 s to diffuse out of
a container. The time required by an equal volume of O2 to diffuse out under identical conditions, is
Magnetic field
An electric field of appropriate magnitude is also applied, so that the electron travels undeviated without any change in its speed through the chamber. We are ignoring gravity. Then, the direction of the electric field is (a) opposite to the direction of the magnetic field (b) opposite to the direction of the electron’s motion (c) normal to the plane of the paper and coming out of the plane of the paper (d) normal to the plane of the paper and into the plane of the paper
(a) 24 s
(b) 96 s
(c) 384 s
(d) 192 s
39. Acetic acid reacts with sodium metal at room temperature to produce (d) CO
(c) H 2O
(b) H 2
(a) CO2
40. The equilibrium constant, kc for 3 C2H2(g) 2
C6H6(g)
−2
is 4 L mol . If the equilibrium concentration of benzene is 0.5 mol L−1, that of acetylene in mol L−1 must be (a) 0.025
(b) 0.25
(c) 0.05
(d) 0.5
41. The weight per cent of sucrose (formula weight = 342
CHEMISTRY 31. The molecule having a formyl group is (a) acetone (c) acetic acid
(b) acetaldehyde (d) acetic anhydride
(a) 0.01
(b) 0.1
(c) 1.0
(d) 10
42. The order of reactivity of K, Mg, Au and Zn with water is
32. The structure of cis3hexene is (a)
g mol −1 ) in an aqueous solution is 3.42. The density of the solution is 1 g mL−1, the concentration of sucrose in the solution in mol L−1 is
(a) K > Zn > Mg > Au (c) K > Au > Mg > Zn
(b)
(b) K > Mg > Zn > Au (d) Au > Zn > K > Mg
43. Which of the following is an anhydride? O (c)
(d)
(a)
33. The number of sp hybridised carbon atoms in O HC
C
CH2
(a) 3
C
(b) 5
CH
(c) 4
CH2, is
(d) 6
34. The number of valence electrons in an atom with electronic configuration 1s22s22 p63s23 p3 is (a) 2
(b) 3
(c) 5
(d) 11
35. The pair of atoms having the same number of neutrons is 23 39 23 24 23 19 24 (a) 12 6 C, 12 Mg (b) 11 Na, 9 F (c) 11 Na, 12 Mg (d) 11 Na, 19 K
CH3
O O
H 3C
CH3 O
O (c)
CH2
O (b) H3C
O
H 3C
2
O
CH3
O
CH3
(d) H3 C
O
44. Which of the following metals will precipitate copper from copper sulphate solution? (a) Hg
(b) Sn
(c) Au
(d) Pt
45. The radii of the first Bohr orbit of H (rH ), He + (rHe + ) and Li 2+ (rLi 2 + ) are in the order
(a) rHe + > rH > r 2 + Li (c) rH > rHe + > r 2 + Li
(b) rH < rHe + < r 2 + Li (d) rHe + < rH > r 2 + Li
97
KVPY Question Paper 2013 Stream : SA
53. Animal cells after removal of nuclei still contained
BIOLOGY
DNA. The source of this DNA is
46. The Bowman’s capsule, a part of the kidney is the site of (a) filtration of blood constituents (b) reabsorption of water and glucose (c) formation of ammonia (d) formation of urea
DNA? (a) Guanine and guanidine (b) Guanidine and cytosine (c) Guanine and cytosine (d) Adenine and guanidine
temperature is controlled by the
(a) Binary fission (c) Budding
medium is (c) bacterium
(d) fungus
49. MeiosisI and meiosisII are characterised by the separation of (a) homologous chromosomes; sister chromatids (b) sister chromatids; homologous chromosomes (c) centromere; telomere (d) telomere; centromere
50. People suffering from albinism cannot synthesise (b) melanin
(c) keratin
(d) collagen
51. Shortsightedness in humans can be corrected by using (a) concave lens (c) cylindrical lens
(b) Multiple fission (d) Conjugation
56. Which one of the following classes of animals
48. A pathogen which cannot be cultured in an artificial
(a) suberin
55. Which one of the following is not a mode of asexual reproduction?
(a) parietal lobe of cerebrum (b) limbic lobe of cerebrum (c) temporal lobe of cerebrum (d) frontal lobe of cerebrum
(b) virus
(b) mitochondria (d) lysosome
54. Which one of the following combinations is found in
47. In human brain, the sensation of touch, pain and
(a) protozoan
(a) nucleosomes (c) peroxisomes
(b) convex lens (d) plain glass
52. A person with blood group ‘A’ can (i) donate blood to and (ii) receive blood from (a) (i) person with blood group ‘AB’ and (ii) persons with any blood group (b) (i) person with blood group ‘A’ or ‘AB’ and (ii) ‘A’ or ‘O’ blood groups (c) (i) person with blood group ‘B’ or ‘AB’ and (ii) ‘B’ or ‘O’ blood groups (d) (i) person with any blood group and (ii) ‘O’ blood group only
constitutes the largest biomass on the earth? (a) Insects (c) Mammals
(b) Fishes (d) Reptilians
57. In the digestive system, the pH of the stomach and the intestine, respectively are (a) alkaline, acidic (c) acidic, neutral
(b) acidic, alkaline (d) acidic, acidic
58. The major nitrogenous excretory product in mammals is (a) amino acids (c) urea
(b) ammonia (d) uric acid
59. Which of the following plant traits (characters) is not an adaptation to dry (xeric) habitats? (a) Sunken stomata on leaves (b) Highly developed root system (c) Thin epidermis without a cuticle on stem and leaves (d) Small leaves and photosynthetic stem
60. Biological diversity increases with the productivity of an ecosystem. In which of the following habitats do we see the greatest diversity of species? (a) Tropical dry grasslands (b) Temperate deciduous forests (c) Alpine grasslands (d) Tropical evergreen forests
PARTII (2 Marks Questions) MATHEMATICS 61. Let a , b, c, d , e be natural numbers in an arithmetic progression such that a + b + c + d + e is the cube of an integer and b + c + d is square of an integer. The least possible value of the number of digits of c is (a) 2 (c) 4
(b) 3 (d) 5
62. On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers so written, there are three distinct numbers and they
are 16, 24 and 31. The volume of the cuboid lies between (a) 7 and 14 (c) 21 and 28
(b) 14 and 21 (d) 28 and 35
63. Let ABCD be a square and let P be a point on
segment CD such that DP : PC = 1 : 2. Let Q be a point on segment AP such that ∠BQP = 90°. Then, the ratio of the area of quadrilateral PQBC to the area of the square ABCD is
(a)
31 60
(b)
37 60
(c)
39 60
(d)
41 60
98
KVPY Question Paper 2013 Stream : SA
64. Suppose the height of a pyramid with a square base is decreased by p% and the lengths of the sides of its square base are increased by p% (where, p > 0). If the volume remains the same, then (a) 50 < p < 55 (c) 60 < p < 65
(b) 55 < p < 60 (d) 65 < p < 70
65. There are three kinds of liquids X , Y , Z . Three jars J1 , J 2 , J3 contains 100 ml of liquids X , Y , Z respectively. By an operation we mean three steps in the following order – stir the liquid in J1 and transfer 10 ml from J1 into J 2, – stir the liquid in J 2 and transfer 10 ml from J 2 into J3 , – stir the liquid in J3 and transfer 10 ml from J3 into J1. After performing the operation four times, let x, y, z be the amounts of X , Y , Z respectively, in J1.Then,
(a) x > y > z (b) x > z > y
(c) y > x > z
(d) z > x > y
PHYSICS 66. Two identical uniform rectangular blocks (with longest side L) and a solid sphere of radius R are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure.
Note That irrespective of speed of P, ball always leaves P’s hand with speed 2 ms −1 with respect to the ground. Ignore gravity. Balls will be received by Q. (a) One every 2.5 s in case (I) and one every 3.3 s in case (II) (b) One every 2 s in case (I) and one every 4 s in case (II) (c) One every 3.3 s in case (I) and one every 2.5 s in case (II) (d) One every 2.5 s in case (I) and one every 2.5 s in case (II)
68. A 10.0 W electrical heater is used to heat a container filled with 0.5 kg of water.It is found that the temperature of the water and the container rose by 3 K in 15 min. The container is then emptied, dried and filled with 2 kg of an oil. It is now observed that the same heater raises the temperature of the containeroil system by 2 K in 20 min. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is (a) 2.5 × 10 3 JK−1 kg −1 (b) 51 . × 10 3 JK−1 kg −1 (c) 3.0 × 10 3 JK−1 kg −1 (d) 15 . × 10 3 JK−1 kg −1
69. A ray of light incident on a transparent sphere at an
angle π / 4 and refracted at an angle r, emerges from the sphere after suffering one internal reflection. The total angle of deviation of the ray is
3π − 4r 2 π (c) − r 4
(a)
π − 4r 2 5π (d) − 4r 2 (b)
70. An electron with an initial speed of 40 . × 106 ms −1 is brought to rest by an electric field. The mass and charge of an electron are 9 × 10−31 kg and 16 . × 10−19 C, respectively. Identify the correct statement.
Table L
R x
If the mass of each block is M and of the sphere is M / 2, then the maximum distance x that can be achieved is (a) 8L / 15 (c) (3L / 4 + R )
(b) 5L / 6 (d) (7L / 15 + R )
67. Two skaters P and Q are skating towards each other. Skater P throws a ball towards Q every 5 s such that it always leaves her hand with speed 2 ms −1 with respect to the ground. Consider two cases: (I) P runs with speed 1 ms −1 towards Q, while Q remains stationary. (II) Q runs with speed 1 ms −1 towards P, while P remains stationary.
(a) The electron moves from a region of lower potential to higher potential through a potential difference of 11.4 µV (b) The electron moves from a region of higher potential to lower potential through a potential difference of 11.4 µV (c) The electron moves from a region of lower potential to higher potential through a potential difference of 45 V (d) The electron moves from a region of higher potential to lower potential through a potential difference of 45 V
CHEMISTRY 71. The degree of dissociation of acetic acid (0.1 mol L−1 ) in water (K a of acetic acid is 10−5 ) is (a) 0.01 (c) 0.1
(b) 0.5 (d) 1.0
99
KVPY Question Paper 2013 Stream : SA 77. A diabetic individual becomes unconscious after
72. Compound X on heating with Zn dust gives compound Y which on treatment with O3 followed by reaction with Zn dust gives propionaldehyde. The structure of X is Br
selfadministering insulin. What should be done immediately to revive the individual? (a) Provide him sugar (b) Give him higher dose of insulin (c) Provide him salt solution (d) Provide him lots of water
Br
(a)
(b) Br
Br
78. A regular check on the unborn baby of a lady towards Br
Br
the end of her pregnancy showed a heart rate of 80 beats per minute. What would the doctor infer about the baby’s heart condition from this?
(d)
(c) Br
Br
(a) Normal heart rate (b) Faster heart rate (c) Slower heart rate (d) Defective brain function
73. The amount of metallic Zn (atomic weight = 65.4) required to react with aqueous sodium hydroxide to produce 1 g of H2 , is (a) 32.7 g
(b) 98.1 g
(c) 65.4 g 12
79. Three uniformly watered plants i, ii and iii were kept
(d) 16.3 g
in 45% relative humidity, 45% relative humidity with blowing wind and 95% relative humidity, respectively. Arrange, these plants in the order (fastest to slowest) in which they will dry up.
13
74. Natural abundances of C and C isotopes of carbon are 99% and 1%, respectively. Assuming they only contribute to the mol. wt. of C2F4, the percentage of C2F4 having a molecular mass of 101 is (a) 1.98
(b) 98
(c) 0.198
(a) i → ii → iii (c) iii → ii → i
(d) 99
75. 2,3dimethylbut2ene when reacted with bromine forms a compound which upon heating with alcoholic KOH produces the following major product.
80. Many populations colonising a new habitat show a logistic population growth pattern over time, as shown in the figure below
(b) Br
OH
Population size
(a)
(b) ii → i → iii (d) iii → i → ii
OH OH
(c)
(d) OH
OH
Time
BIOLOGY
In such a population, the population growth rate
76. Sister chromatids of a chromosome have
(a) stays constant over time (b) increases and then reaches a asymptote (c) decreases over time (d) increases to a maximum and then decreases
(a) different genes at the same locus (b) different alleles of the same gene at the same locus (c) same alleles of the same gene at the same locus (d) same alleles at different loci
Answers PARTI 1 11 21 31 41 51
(c) (b) (b) (b) (b) (a)
2 12 22 32 42 52
(c) (c) (d) (c) (b) (b)
3 13 23 33 43 53
(c) (a) (b) (a) (a) (b)
4 14 24 34 44 54
(b) (b) (a) (b) (b) (c)
5 15 25 35 45 55
(c) (b) (c) (c) (c) (d)
6 16 26 36 46 56
(a) (a) (c) (d) (a) (a)
7 17 27 37 47 57
(c) (b) (a) (d) (a) (b)
8 18 28 38 48 58
(c) (a) (b) (b) (b) (c)
9 19 29 39 49 59
(c) (d) (a) (b) (a) (c)
10 20 30 40 50 60
(d) (c) (c) (d) (b) (d)
62 72
(d) (c)
63 73
(d) (a)
64 74
(c) (a)
65 75
(b) (b)
66 76
(a) (c)
67 77
(a) (a)
68 78
(a) (c)
69 79
(a) (b)
70 80
(d) (d)
PARTII 61 71
(b) (a)
100
KVPY Question Paper 2013 Stream : SA
Solutions 1. (c) We have, x + y + z = 10
4. (b) We have,
Let three number x + 1, y + 1, z + 1
⇒ ⇒ ⇒
AM ≥ GM (x + 1) + ( y + 1) + (z + 1) ≥ 3 [(x + 1) ( y + 1) (z + 1)]1/3 x+ y+ z+ 3 ≥ ⇒ 3 (xyz + xy + yz + xz + x + y + z + 1)1/3 3
13 ≥ xyz + xy + yz + xz + 11 3
⇒
Now, x, y, z are integer. ∴ xyz + xy + yz + xz + 11is also integer. 3
13 ∴ is also integer. 3 ∴
13 3 13 3 . = 81 Q = 8137 3 3
∴xyz + xy + yz + xz + 11 ≤ 81 ⇒
2. (c) Given, 2013 + a = b 2
⇒
b ,−1 a It has at least one negative root, i.e. − 1. So, it has both roots are real. ∴Option (b) is correct. ⇒
x=−
5. (c) Given, ⇒
x y z y z x + + = 7⇒ + + = 9 y z x x y z
We know that, a3 + b3 + c3 − 3abc = (a + b + c) (a 2 + b2 + c2 − ab − bc − ca ) ∴a3 + b3 + c3 − 3abc = [(a + b + c)2 − 3(ab + bc + ca ))] 3
3
2
b2 − a 2 = 2013 (b − a ) (b + a ) = 3 × 11 × 61
ab is minimum. When b − a = 33 and b + a = 61 On solving, we get a = 14 and b = 47 ∴Minimum value of ab = 14 × 47 = 658
3. (c) We have sides of triangle are, b + 5, 3b − 2, 6 − b Triangle are isosceles. ∴Two sides are equal.
8. (c) Given, ABC is an acute angle triangle. ∠B = 45° A C1
B1
3
2 x x y z y z + + − 3 + + z x x y y z
x3
∴
3
y
+
y3 3
z
+
45°
B
− 3 = (7) [72 − 3 × 9] x3 = 7 (49 − 27) = 7 × 22 = 154 B D
A
C
In ∆ABD, ∠D > ∠C So, ∆ABD not similar to ∆ABC.
7. (c) Given, PR and QR are tangents.
A1
∠ADC = 90° [Q AD is altitude] ∠BAD = 45° = ∠BAA ′ Similarly, ∠BCC1 = 45° ∴ ∠BAA1 = ∠BB1 A1 [Q angle on same segment are equal] ∠BCC1 = ∠BB1C1 [Q angle on same segment are equal] ∴ ∠A1 B1C1 = ∠BB1 A1 + ∠BB1C1 = 45° + 45° = 90° 9. (c) Given, ABCD is rectangle. ∴ AB = CD, BC = AD
P 5 R
Case II 3b − 2 = 6 − b ⇒ b = 2
r
4
6 M r Q
1 2
∴
2x
A
B
y
3
5
∴Sides are 7, 4, 4
11 1 11 Sides are , − , which is not 2 2 2 possible. ∴ Only for two values of b, triangles are isosceles.
θ
C
D
z3
6. (a) In ∆ABC, ∠A < ∠B < ∠C
Case I b + 5 = 3b − 2 7 b= ∴ 2 17 17 5 So, sides are , , . 2 2 2
Case III b + 5 = 6 − b ⇒ b =
In ∆PRM, RM 2 = PR 2 − PM 2 = 25 − 9 = 16 RM = 4 PM 3 …(i) In ∆PRM, tanθ = = RM 4 OP r …(ii) In ∆POR, tanθ = = PR 5 From Eqs. (i) and (ii), we get 15 3 r = ⇒r = 4 4 5
x x y z y z ⇒ + + − 3 = + + x z y z y x
xyz + xy + yz + xz ≤ 70
∴Maximum value of xyz + xy + yz + xz is 69. ⇒
ax2 + (a + b)x + b = 0 ax2 + ax + bx + b = 0 (ax + b) (x + 1) = 0
PQ = 6 PR = 5 1 PM = PQ 2 1 PM = × 6 = 3 2
O
2y
X y E
2x
D
x
Y
x
C
2y
F
X and Y are midpoint of AD and CD respectively.
101
KVPY Question Paper 2013 Stream : SA Let AB = 2x , BC = 2 y ∴AX = XD = y DY = YC = x Area of rectangle ABCD = 4xy = 60 ⇒ xy =15 In ∆ABX and ∆DEX, ~ ∆DEX ∆ABX − ∴ DE = AB = 2x Similarly, ∆CBY ~ − ∆DFY
1 × AQ × AP × sin 30° Area of ∆PAQ 2 = ∴ 1 Area of ∆RSP × RS × PS × sin 30° 2 2× 2 = =2 1 [Q AQ = AP = 2 , RS = PS = 1]
11. (b) Given, X complete one round in 40 s. ∴ 2 π = 40 s
∴FD = BC = 2 y ∴Area of ∆BEF = Area of ∆EFX + Area of ∆ BFX 1 1 = FX ⋅ DE + FX ⋅ AB 2 2 1 1 = × 3 y × 2x + × 3 y × 2x 2 2 = 6xy [Qxy =15] = 6 × 15 = 90
10. (d) Given, A
1
S
1
1
F
R 1
Q
1
1
1
P 1
E
C
D
ABCDEF is a regular hexagon of side length 1. ABQR and AFPS is a square of each side length also 1. ADCDEF is a regular hexagon ∴
∠FAB = 120°
In square ABQR, AB = BQ = 1 AQ is a diagonal of square ∴
AQ =
AB 2 + BQ 2 =
2
⇒ ∠BAS = ∠FAB − ∠FAS
= 120° − 90° = 30° ⇒ ∠SAR = ∠BAR − ∠BAS
= 90° − 30° = 60° ⇒ ∠ASR = 60°
[Q ∆ARS is an equilateral triangle] ⇒ ∠RSP = ∠ASP − ∠ASR
= 90° − 60° = 30° ⇒ ∠FAB = ∠FAP + ∠PAQ + ∠QAB
⇒ 120° = 45° + ∠PAQ + 45° [Q ∠FAP = ∠QAB = 45° FA = FP and AB = BQ] ∴ ∠PAQ = 30°
Put n = 1, 2, 4, 5 n! + 10 is not a perfect square. Put n = 3, 3! + 10 = 6 + 10 = 16 is a perfect square. If n>5 n! is multiple of 10. ∴ n ! = 10k n ! + 10 = 10k + 10 = 10(k + 1) (when, k is even) = 10 (2m + 1) = 2 × 5 (2m + 1) ∴Product of odd and even is not a perfect square.
14. (b) We have 10 points lie a plane such that no three of them are collinear. 2
2π In one second, he complete, round 40 2 π × 15 In 15 s, he complete round 40
B
1
θ
13. (a) Given, n! + 10
LetY complete one round in t s ∴ 2π = t 2π In one second Y complete round t 2π In 15 s,Y complete × 15 round t Since, both are move in opposite direction. 2π 2π × 15 + × 15 = 2 π ∴ 40 t 1 1 15 + = 1 ⇒ 40 t 1 1 1 = − ⇒ t 15 40 1 8− 3 ⇒ = t 120 5 1 = = 120 24 ∴ t = 24 s
12. (c) We have, n + 1 − n − 1 < 0.2, n ∈ N n + 1 < 0.2 + n − 1 ⇒ On squaring both side, we get n + 1 < 0.04 + n − 1 + 0.4 n − 1 ⇒ n + 1 − n + 1 − 0.04 < 0.4 n − 1 2 − 0.04 ⇒ < n −1 0.4 ⇒ 4.9 < n − 1 ⇒ n − 1 > (4.9)2 ⇒ n > 1 + 24.01 ⇒
n > 25.01
∴Minimum value of n = 26
3 4 5
1
6 10
9
8
7
According to question only 5 ways are possible i.e. 1 6, 2  7, 3  8, 4  9 and 5  10.
15. (b) Let total number of people whose salary less than ` 10000 per annum = x and annual salary of each person = a ∴Total salary = ax and total number of people whose salary more than ` 10000 per annum = y and annual salary of each person = b ∴Total salary = bx When 5% increase of salary of people x 105ax i.e. x (a + 5% of a) = 100 and 5% decrease of salary of people y 95 by i.e. y (b − 5% of b) = 100 105ax 95by + Average salary after 100 = 100 Average salary before ax + by = 1+
5 ax − by 100 ax + by
ax − by < 0 ∴Average salary after be decreases.
16. (a) As centre of mass is subjected to a downward external force only, so its motion is along the direction of external force, i.e. downwards. Any internal force does not change position of centre of mass.
102
KVPY Question Paper 2013 Stream : SA
17. (b) When ball is released, vertical component of ball’s velocity first increases in negative direction (downwards), then on collision with floor, its velocity is reversed (upwards). As acceleration remains constant, so lines are parallel to each other as given in option (b).
18. (a) Pressure of a fluid column depends only on height of fluid column and as pressure is scalar, its magnitude does not depend on orientation of surface over which pressure acts.
19. (d) We have following observations from position x versus time t graph. Velocity or slope is again negative
x
Slope or velocity is negative Slope or velocity x is positive decreases Slope or velocity changes from negative to positive
t
From above graph we can draw following velocity v versus time t graph. v
As both g and ∆t are constants. ∴∆v is constant and its value does not changes with time t. The mutual separation ∆s of the stones is ∆s = s1 − s2 1 1 = − gt 2 − − g (t − ∆t )2 2 2 1 = g ((t − ∆t )2 − t 2 ) 2 1 = g (t 2 + ∆t 2 − 2t∆t − t 2 ) 2 1 = g (− 2t∆t + ∆t 2 ) 2 1 ⇒ ∆s = g (∆t 2 − 2t∆t ) 2 Clearly, ∆s decreases with time and becomes zero when 2t = ∆t.
21. (b) Refractive index of a material is inversely proportional to wavelength of light. 1 µ∝ ⇒ λ Now, λ yellow > λgreen > λ blue ⇒ µ yellow < µ green < µ blue or µ3 < µ1 < µ 2 22. (d) As focal length of lens is 20 cm, point of
Slope or acceleration is positive
I
Slope or acceleration is negative I′
M
P
t
So, acceleration of given particle is initially positive but with time it becomes negative. Hence, a = p − qt is best suited option.
20. (c) Let first stone mass m1 is dropped at instant t = 0. Then at time t, its velocity and displacement respectively, are 1 v1 = − gt and s1 = − gt 2 2 As, second stone mass m2 is dropped ∆t time after, so its velocity and displacement at instant t respectively, are v2 = − g (t − ∆t ) 1 and s2 = − ( g ) (t − ∆t )2 2 Difference in speeds of stones is ∆v = v1 − v2 = (− gt ) − (− g (t − ∆t )) = − gt + gt − g∆t = − g∆t
convergence of a parallel beam of light is also 20 cm. Now, given PM = 10 cm So, PI ′ = PM + MI = 20 cm or MI = 20 − 10 = 10 cm As, PI = 10 cm ∴∆PMI is an equilateral triangle of side 10 cm. Now, if MN is normal to mirror, as angle of incidence and reflection are equal, we have following situation I 30° N 30° P
α
α M
I′
From above figure, we have α + 30° = 90° ⇒α = 60° Hence, mirror makes an angle of 60° with the horizontal.
23. (b) Rear view mirror is a convex mirror. Here, u = − 10 m 1.5 R f = = + m ∴ 2 2 Now, from mirror equation, 1 1 1 1 1 1 2 1 + = or = − = − v u f v f u 1.5 (−10) 4 1 43 30 or v = m = + = 3 10 30 43 Now, magnification, 30 − 43 −v = m= = 0.069 or m = 0.07 u − 10
24. (a) first we distribute current in circuit given as I2–I4 I2 I3
2 1
I4
I5 I2–I4 –I5
3 4
I1 I2+I3
Current distribution must follows Kirchhoff’s junction rule. Now, from closed loops marked 1, 2, 3 and 4, we have following set of equations by application of Kirchhoff’s loop rule, …(i) I1 = I 2 + I3 …(ii) I3 = I 2 + I 4 I 4 = I 2 − I 4 + I5 …(iii) ⇒ 2I 4 = I 2 + I5 I5 = 2 (I 2 − I 4 − I5 ) …(iv) ⇒ I5 = 2I 2 − 2I 4 − 2I5 …(v) 3I5 = 2I 2 − 2I 4 From Eqs. (iii) and (v), we have 3I5 = 2I 2 − (I 2 + I5 ) …(vi) ⇒ 4I5 = I 2 From Eqs. (iii) and (vi), we have 5 …(vii) 2I 4 = 4I5 + I5 ⇒ I 4 = I5 2 From Eqs. (ii), (vi) and (vii), we have 5 13 …(viii) I3 = 4I5 + I5 = I5 2 2 Now, marked currents I and I ′ in the given circuit are 5 I ′ = (I 2 − I 4 − I5 ) = 4I5 − I5 − I5 2 I 8 − 5 − 2 …(ix) = I5 = 5 2 2 And I = I 2 = 4I5 Hence, ratio of I / I ′ = (4I5 ) / (I5 / 2) = 8.
103
KVPY Question Paper 2013 Stream : SA 25. (c) An emf is induced in the coil when there is a flux change in the coil or when field lines are cut by the coil. When coil rotates about its axis, there is no change in flux as no field line is cut by the coil. ω
=
. × 103 × (15 . × 1011 )2 4 × 22 × 14 7 × (3 × 108 )2
≈ 109 kg s−1 So, mass reduction per second is around 109 kg.
28. (b) Power dissipated by resistor is P = I 2R ⇒
B
So, no emf is generated in coil when it is rotated about its axis. In all other cases an emf is induced in the coil.
26. (c) Given circuit is A 0.1A
80 Ω V
29. (a) Initially let pressure is p, then 20 Ω
– 60 Ω
20 Ω B
20 × 60 = 15 Ω 20 + 60 We can redraw the circuit as Req =
I2
A 0.1A
80 Ω V
2∆I ∆P × 100 = × 100 P I Per cent change in power dissipation = 2 × Per cent change in current = 2 × 3% = 6%
25 Ω
+
20 Ω
– 15 Ω B
Let current through 20 Ω resistor is I 2, then V AB = 01 . (25 + 15) = I 2 × 20 01 . × 40 or I2 = = 0.2 A 20 So, by Kirchhoff’s junction rule, current through 8 Ω resistance is I = 01 . + 0.2 = 0.3 A.
pV = n1 RT V . 3 Let number of moles of gas left is n2, then V p = n2RT 3 Dividing both equations, we get n n2 1 n − n2 2 1 or 1 − 2 = 1 − or 1 = = 3 n1 n1 3 n1 3 n1 − n2 2 Hence, × 100 = × 100 = 66% 3 n1 or percentage of number of molecules escaped = 66%.
30. (c) Using right hand rule, direction of magnetic force can be found. It acts outward from the plane of paper in given case. ∧
r Sun
(i) Acetone
e
O
v∧ i
∧
Fm
To avoid deflection of electron, electric field must be applied normal to the plane of paper (XOY plane in diagrams) and pointing outward.
C
CH3
O (ii) Acetaldehyde
CH3
(iii) Acetic acid
H
CH3OH
(iv) Acetic anhydride CH3
C
O
C
O
CH3
O
As acetaldehyde has R C H group,  O Thus, the correct option is (b).
32. (c) The structure of cis3hexene is 2
5
1 3
.
6
4
H
H
33. (a) Generally a sp 2hybridised carbon atom is one which has a double bond. O HC
C
sp
sp
CH2 sp3
C sp2
CH2 sp3
CH sp2
CH2 sp2
34. (b) Valence electrons are those electrons which are present in the outermost shell of an element. In the given electronic configuration, 1s2 2s2 2 p6 3s2 3 p3 , 3p is the outermost shell, thus the number of valence electrons are 3. 35. (c) Number of neutrons in the pairs given in options are as follows 12 6
C,
24 12
Mg
No. of neutrons in C = 12 − 6 = 6 No. of neutrons in Mg = 24 − 12 = 12
B
Fe
O
C O
(a)
y
So, total energy radiated per second from sun is ∆E = 4 πr 2 × 14 . × 103 22 . × 103 × (15 . × 1011 )2 = 4× × 14 7 From E = mc2 , we have ∆E ⇒ ∆m = 2 c
CH3
Thus, the given structure has 3 sp 2hybridised carbon atoms.
j B
–k
Earth
O
The structure of the given organic compounds are as follows
Finally pressure is p and volume is
27. (a) Energy from sun is radiated in a sphere of radius (r = 15 . × 1011 m).
R — C — H group. 
or
25 Ω
+
∆P 2∆I = P I
31. (b) A formyl group is one which consists of a carbonyl group attached to a hydrogen, i.e.,
E
(b)
Magnitude of electric field applied is such that Fm = Fe .
Na ,
11
19 9
F
No. of neutrons in Na = 23 − 11 = 12 No. of neutrons in F = 19 − 9 = 10
x v
Fm
23
(c)
23
Na ,
11
24 Mg 12
No. of neutrons in Na = 23 − 11 = 12 No. of neutrons in Mg = 24 − 12 = 12
104 (d)
23
KVPY Question Paper 2013 Stream : SA
Na ,
11
39 K 19
40. (d) For the reaction,
No. of neutrons in Na = 23 − 11 = 12 No. of neutrons in K = 39 − 19 = 20 Thus, option (c) is correct.
36. (d) A molecule which has a symmetrical geometry will have no dipole moment, as the magnitude of all the bond moments cancel each other. The structure of compounds given in options are as follows Cl
C
, Cl
H H CH3Cl (µ = 0)
H
,
C Cl
Cl Cl CHCl3 (µ = 0) Cl
H
C
Cl
Cl
Kc = [C2H2 ]3 = [C2H2 ] =
[C6 H6 ] [C2H2 ]3
⇒4=
0.5 [C2H2 ]3
0.5 1 = 4 8 1 = 0.5 mol/L 2
Li
46. (a) In Bowman’s capsule
41. (b) Given, 3.42 = 0.0342 g 100 Molar mass of sucrose = 342 g Mass of solution = 100 g Density of solution = 1 g mL−1 0.0342 No. of moles of sucrose = ∴ 342 Mass Also, density = Volume 100 Volume = = 100 mL ∴ 1 Concentration of solution is calculated in terms of molarity No. of moles of sucrose Molarity = Vol. of solution in litres 0.0342 0.0342 . mol/L = = × 1000 = 01 100 100 1000 Weight of sucrose =
H
C H
3C2H2 ( g ) r C6 H6 ( g )
Cl Cl
CCl4 Symmetrical molecule, µ =0
Thus, CCl 4 has no dipole moment.
37. (d) As the species are radioactive, so they follow Ist order kinetics. For Ist order Ct = C0 e− kt 1 k∝ ∴ t From the graph it can be concluded that species A takes maximum time to decay while species C takes least time. Thus, decay constant follows the order kC > kB > kA .
42. (b) The reactivity of K, Mg, Au and Zn with water can be determined by the reactivity series of metals. According to reactivity series, the decreasing order of reactivity will be K > Mg > Zn > Au
38. (b) The ratio of rate of diffusion of
43. (a) An anhydride is a compound that
two gases, O 2 and H 2 can be given as rO 2 MH 2 = rH 2 MO 2
has two acyl groups bonded to same oxygen atom, i.e.
rO 2 rH 2
=
2 1 = 32 4
R
C
O
R
C
O
O
This type of structure is given in option (a).
rO 2 : rH 2 = 1 : 4
O
Also, rate of diffusion Volume of diffused gas = Time of diffused gas rO 2 tH 2 1 24 = = = 96 s ∴ = rH 2 tO 2 4 tO 2
39. (b) Whenever an acid reacts with a metal, hydrogen gas is evolved. So, when acetic acid reacts with a sodium metal, hydrogen gas is produced. −
CH3COOH + Na → CH3COONa + + H2 ↑
CH3
45. (c) According to Bohr’s radius of an atom n2 1 ⇒ r∝ r = 0.529 × Z Z where, Z is the atomic number. Thus, more is the atomic number lesser will the Bohr’s radius. Therefore, the correct order is rH > rHe + > r 2 +
O O
CH3
2 acyl groups attached to Oatoms
Thus, option (a) is correct.
44. (b) More reactive metal than Cu can precipitate copper from copper sulphate solution. The increasing order of reactivity of given element is Au < Hg < Cu < Sn As Sn is more reactive than Cu, so it would precipitate Cu from CuSO4 .
ultrafiltration of blood occurs. Bowman’s capsule is a cuplike sack at the beginning of the tubular component of a nephron in the mammalian kidney that performs the first step in the filtration of blood to form urine. Fluids from blood in the glomerular are collected in the Bowman’s capsule (i.e. glomerular filtrate) and further processed along the nephron to form urine. This process is known as ultrafiltration.
47. (a) Parietal lobe is sensory lobe for touch, pain and temperature. The cerebral cortex is divided into four sections, called ‘lobes’. Out of these, the parietal lobe is associated with movement, orientation, recognition and perception of stimuli. Thus, the parietal lobe functions in registration of sensory perception of touch, pain, heat and cold, knowledge about position in space, taking in information from environment, organising it and communicating to rest of brain.
48. (b) Virus cannot be cultured in an artificial medium. It multiples only in living cells. Viruses are obligate intracellular parasites. They lack metabolic machinery to generate energy or to synthesise proteins, instead they rely on their host cells to carry out these functions.
49. (a) MeiosisI Reduction division (2n → n ), separation of homologous chromosomes results in reduction of chromosome ploidy to half. MeiosisII Similar to mitosis where sister chromatids separate. 50. (b) Melanin pigment synthesised from tyrosine amino acid, imparts colour to skin. People suffering from albinism cannot synthesise melanin. Albinism is a disease, in which a person has partial or complete loss of pigmentation (colouring) of the skin, eyes and hair. There is a cell called the melanocyte that is responsible for giving eyes, skin and hair pigmentation.
105
KVPY Question Paper 2013 Stream : SA In albinism, there occurs genetic mutation in melanocytes which interfere with their pigment.
51. (a) Shortsightedness (myopia) is corrected by using concave lens. These lens work by bending the light rays slightly outwards, so that they can focus further back on the retina. Myopia is an eye defect in which the eyeball grows slightly too long. This means that light does not focus on the light sensitive tissue (retina) at the back of the eye properly. Instead, the light rays focus just in front of the retina, resulting in distant objects appearing blurred.
52. (b) A person with blood group ‘O’ is a universal donor, whereas person with blood group ‘AB’ is a universal recipient. Blood group ‘O’ do not have any antigen on their RBC whereas blood group ‘AB’ do not have any antibody in their blood plasma, but both the antigens A and B on their RBC. Therefore, a person with blood group ‘A’ can donate blood to a person with blood group ‘A’ or ‘AB’ and can receive blood from a person with blood group ‘A’ or ‘O’.
53. (b) Mitochondria are structures within cells that convert the energy from food into a form that cells can use. Although most DNA is packaged in chromosomes within the nucleus, mitochondria also have a small amount of their own DNA. This genetic material is known as mitochondrial DNA or mtDNA. Thus, after the removal of nuclei, the cell still have mtDNA. 54. (c) The correct combination present in DNA is guanine and cytosine. The DNA consists of four types of nitrogen bases, i.e. adenine (A), thymine (T), guanine (G) and cytosine (C). Whereas guanidine is a strong base that found in urine as a normal product of protein metabolism and not present in DNA. 55. (d) Conjugation is the transfer of genetic material between bacterial cells by direct cell to cell contact or by a bridgelike connection between two cells (e.g. bacteria). Thus, conjugation is a process of genetic recombination not asexual reproduction. 56. (a) Insects (class–Insecta or Hexapoda) are the animals constituting the largest biomass on the earth. In the world, about 900 thousand different kinds of living insects are known. This represents approximately 80% of the world’s animal species.
57. (b) In the digestive system, the pH of stomach and intestine are acidic and alkaline, respectively. The pH of stomach is 1.52.5 (i.e. acidic) and the pH of intestine is 7.47.6 (i.e. alkaline). 58. (c) The major nitrogenous excretory product in mammals is urea. Nitrogenous wastes in the body of animals tend to form toxic NH 3 , which must be excreted. NH 3 is converted to urea in hepatocytes of the body. 59. (c) Thin epidermis without a cuticle on stem and leaves is not an adaptation to dry habitat. This is because the cuticle is a waxy layer on the epidermis which prevents the entire leaf from losing water from the surface. Thus, thick cuticle prevents water loss. 60. (d) Diversity of species is highest in the tropical evergreen forests primarily because there are fewer ecological obstacles for biodiversity. Like the climate is wet and warm, plants and animals have the greatest access to consistent energy, water and carbon, etc. This reduces the selection for traits that emphasise the ability to withstand environmental stresses such as cold and drought, etc., and promotes higher rates of speciation.
62. (d) Let the length, breadth and height of cuboid be x, y and z respectively. ∴Perimeter of face PQRS = 2(x + y) R
S y P
Q
x z
D
C
y A
B
x
Area of PQRS = xy …(i) ∴ 2 (x + y) + xy = 16 Similarly, for face APSD, …(ii) 2( y + z ) + yz = 24 and for face APQB, …(iii) 2(x + z ) + xz = 31 From Eqs. (ii) and (iii), we get …(iv) (x − y ) (2 + z ) = 7 From Eqs. (ii) and (iv), we get …(v) 4x = 2 + 5 y On solving Eqs. (i) and (v), we get x = 3, y = 2, z = 5 ∴Volume of cuboid = xyz = 3 × 2 × 5 = 30 Hence, option (d) is correct.
63. (d) Given, ABCD is a square. D
61. (b) We have, a , b, c, d , e are natural number and in AP. Let D is common difference of AP. ∴Let c=C
x
P
2x
C
Q 3x
3x
a = C − 2D b=C − D d =C + D e = C + 2D a + b + c + d + e = 5C and b + c + d = 3C Given, a + b + c + d + e is a cube of number …(i) ∴ 5C = λ3 and b + c + d is a square of number …(ii) ∴ 3C = u 2 From Eqs. (i) and (ii), we get λ3 u2 = 5 3 λ3 and u 2 is a multiple of 15. ∴Smallest possible value of λ = 15 and u = 45 u 2 (45)2 ∴ c= = = 675 3 3 ∴Number of digits = 3
A
B
3x
AB = BC = CD = AD = 3x PD : PC = 1 : 2 ∴ PD = x PC = 2x In ∆DAP and ∆QBA, ∠DAP = ∠QBA ∠D = ∠Q = 90° ∴ ∆DAP ~ ∆QBA Let
DA AP DP = = QB BA QB
∴ ⇒
10 x 3x x = = QB 3x QA [Q AP =
∴
QB =
9x2 + x2 = 10 x]
9 3 x ⇒ QA = x 10 10
106
KVPY Question Paper 2013 Stream : SA
Area of quadrilateral BQPC = area of square ABCD − (area of ∆APD + area of ∆ABQ) 1 1 9 3 = (3x)2 − × 3x × x + × x× x 2 2 10 10 3 27 2 123 x2 = 9x − x2 + x = 2 20 20 2
123x2 Area of quadrilateral PQBC = 202 Area of square ABCD 9x 41 = 60
64. (c) Let the side of square base of pyramid is x m and height of pyramid is y m.
y
65.(b) We have, three kind of liquids x, y, z and three jars J1 , J 2 , J3 contains 100 ml of liquids X , Y , Z respectively. When 10 ml of J1 transfer to J 2 ∴J1 = 90 ml of X, J 2 = 100 ml ofY and 10 ml of X. When 10 ml of J 2 transfer to J3 1000 100 ofY and of X, J3 = 100 ml J2 = 11 11 10 100 of X of Z1 , ofY and 11 11 When 10 ml of J3 transfer to J1 1100 1100 110 of Z1 , ofY1 , of X and J3 = 11 11 11 2
1 100 ofY and J1 = 90 + 10 × of X1 , 11 121 100 of Z 11 Similarly, we can find four operation of amount of X , Y , Z in J1 . We get x > z > y.
66. (a) For system to be in equilibrium x
without toppling, following conditions must be fulfilled. (i) Centre of mass C1 of sphere and upper block must lie inside the edge of lower block.
x
L–y 2
x
Volume of pyramid 1 1 = area of base × height = x2 y 3 3 When x is increased by p%, then new length= x + p% of 100 + p x = x 100 When y is decreased by p%, then new height 100 − p = y − p% of y = y 100 Now, volume is same. 2
∴
1 2 1 100 + p 100 − p x y = x y 100 3 3 100
⇒
100 + p 100 − p 1 = 100 100
C y
Taking origin of axes choosen at C, we have M L × y = M − y 2 2 y L L or y = + y= ⇒ 2 2 3 (ii) Centre of mass of both of block and sphere must lie inside the edge of table.
3x L L L − + x− − =0 2 2 3 2 5x 4L = 2 3 8L x= 15
⇒ ⇒ ⇒
67. (a) Case I P runs towards Q, while Q is stationary. P
Q
First ball is received at time t1 =
∴Next ball is received at time t2 x−5 x 5 = + 5= + 2 2 2 5 So, ∆t = t2 − t1 = s = 2.5 s 2 Case II Q runs towards P, while P is stationary. P
Q
x First ball is received at time t1 = . 3 ∴Next ball is received at time t2 x−5 = + 5 3 x 10 = + 3 3 10 So, s = 3.3 s ∆t = t2 − t1 = 3
68. (a) Energy supplied by heater = Heat absorbed by water + Heat absorbed by oil So, with water in container, P∆t = mw sw ∆T + mo so ∆T ⇒10 × 15 × 60 = 0.5 × 4200 × 3 + mo so × 3 ⇒ mo so = 900 J K −1 Now with oil in container, P∆t = mo so ∆T + mc sc ∆T ⇒ 10 × 20 × 60 = 2 × so × 2 + 900 × 2 10200 = 2.5 × 103 J K−1 kg −1 ⇒ so = 4 69. (a) According to condition given in question, ray diagram of sphere is
2
⇒ (100) (100) = (10000 + 200 p + p ) (100 − p ) 2 2 ⇒ p + 100 p − 100 = 0 ⇒ p 2 + 100 p + (50)2 = (100)2 + (50)2 ⇒ ( p + 50)2 = 12500 p + 50 = 12500 = 11180 . ⇒ ⇒ p = 11180 . − 50 ⇒ p = 6180 . ∴ 60 < p < 65 2
2
C1 C2 L/2
r r C
L/3 R
x
So, again taking centre of mass C2 as origin, L L L 3M x − + M x − − = 0 2 3 3 2
x . 2
π/4
π/4
So, deviations are δ1 =
π −r 4
107
KVPY Question Paper 2013 Stream : SA
where, V is stopping potential. mv2 2q
V =
9 × 10−31 × (4 × 106 )2
=
2 × 16 . × 10−19
≈ 45 V So, electron must move across a potential difference of 45 V from higher to lower potential.
71. (a) Given, Concentration of acetic acid, C = 01 . M K a of acetic acid = 10−5 According to Ostwald dilution law, Ka = α 2C 10−5 = α 2 × 01 . 10−5 −5 = 10 × 10 = α 2 01 . α 2 = 10−4 α = 10−2
72. (c) Skeletal diagram of given information can be drawn X
Zn
Y
O3/Zn, H2O
2CH3CH2CHO
73. (a) Zn + 2NaOH(aq) → Na 2ZnO4 + H2 ↑ 1 mole of H2 is produced by 1 mole of Zn, i.e. 2 g of H2 is produced by 65.4 g of Zn 65.4 g = 32.7 g ∴1 g of H2 is produced by = 2 74. (a) In molecular formula of C2F4 , there are 4 F atoms. F has atomic mass of 19, so 4 F would have atomic mass of 76 g. So, possible molar mass of C2F4 are 100 (76 + 12 + 12), 102 (76 + 13 + 13) or 101 (76 + 12 + 13). Now according to % abundance % of C2F4 of molar mass 100 (when both the C are 12 C) 1 1 = × × 100 = 0.01% 100 100 % of C2F4 of molar mass 102 (when both the C are 13 C) 99 99 = × × 100 = 98.01% 100 100 % of C2F4 of molar mass 101 = 100 − (98.01 + 0.01) = 198 . % 75. (b) 2,3 dimethylbut2ene when reacts with bromine forms 2,3 dibromo 2,3 dimethyl butane which upon heating with alcoholic KOH produces 2,3 dimethylbut1,3diene as a major product. The reaction can be written as Br
Br
Br2/CCl4 2, 3 dimethyl but2ene
2, 3 dibromo 2, 3 dimethyl butane alc. KOH
Propionaldehyde
2, 3 dimethylbut1, 3diene
On retro synthesis. O
Zn, H2O Ozonolysis
O
2CH3CH2CHO
O
Ozonide Ozonolysis O3
(Y) ∆ Zn dust
Br
Br (X)
76. (c) Sister chromatids contain the same allele of the same gene at the same loci whereas nonsister chromatids contain different alleles of the same gene in the same loci. Sister chromatids are the two chromatids of a replicated chromosome, which are connected by the centromere. They are identical to each other since they are produced by DNA replication. 77. (a) Insulin lowers blood sugar level and in this case, brain is getting inadequate sugar/glucose. Therefore, the person becomes unconscious. In order to
revive the individual we need to provide him sugar, so that the blood sugar level becomes normal.
78. (c) A normal Foetal Heart Rate (FHR) usually ranges from 120160 beats per minute. It is measurable sonographically. Therefore in this case, where the foetal heart rate is 80 beats per minute is a slower heart rate (foetal bradycardia).
79. (b) The plants in the order (fastest to slowest) in which they will dry up is ii → i → iii. Relative humidity is the amount of water vapour present in air expressed as a percentage of the amount needed for saturation at the same temperature. As plants transpire, the humidity around saturates leaves with water vapour. When relative humidity levels are too high or there is a lack of air circulation, a plant cannot make water evaporate by transpiration or draw nutrients from the soil. Therefore 95% relative humidity will dry up the slowest, and the 45% relative humidity with blowing wind will dry up the fastest.
80. (d) The population growth pattern shown here is Sshaped (Sigmoidal curve). It is a population growth curve that shows an initial rapid growth (exponential growth) and then it slows down (decreases) as the carrying capacity is reached. Carrying capacity is the maximum number of individuals in a population that the environment can support. Carrying capacity
Growth rate decreasing Population size
δ 2 = π − 2r π δ3 = − r 4 Total deviation of light ray is 3π − 4r δ = δ1 + δ 2 + δ3 = 2 1 70. (d) From, qV = mv2 2
Growth rate increasing
Time
KVPY Question Paper 2012 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2012 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
l
PARTI (1 Mark Questions) MATHEMATICS 1. Let f (x) be a quadratic polynomial with f (2) = 10 and f (− 2) = − 2. Then, the coefficient of x in f (x) is (a) 1
(b) 2
2. The square root of (a) 1
(b) 2
(c) 3
(d) 4
(0.75)3 + [0.75 + (0.75)2 + 1] is 1 − (0.75) (c) 3
(d) 4
3. The sides of a triangle are distinct positive integers in an arithmetic progression. If the smallest side is 10, the number of such triangles is (a) 8 (c) 10
(b) 9 (d) infinitely many
4. If a , b, c, d are positive real numbers such that
a a + b a + b+ c a + b+ c+ d a , then = = = 3 4 5 6 b + 2c + 3d is
1 (a) 2 (c) 2
22 + 42 + 62 + ... + (2n )2
5. For
(b) 1 (d) not determinable
12 + 32 + 52 + ... + (2n − 1)2 maximum value of n is
(a) 99
(b) 100
to exceed 1.01, the
(c) 101
(d) 150
6. In ∆ABC, let AD, BE and CF be the internal angle bisectors with D, E and F on the sides BC , CA and AB respectively. Suppose AD, BE and CF concur at I and B, D, I, F are concyclic, then ∠IFD has measure (a) 15° (b) 30° (c) 45° (d) any value ≤ 90°
7. A regular octagon is formed by cutting congruent isosceles right angled triangles from the corners of a square. If the square has side length 1, the side length of the octagon is (a) (c)
2−1 2 5−1 4
(b) 2 − 1 (d)
5−1 3
109
KVPY Question Paper 2012 Stream : SA 8. A circle is drawn in a sector of a larger circle of radius r, as shown in the figure given below.
1 h more to 2 complete the work than if both worked together. How much time would they take to complete the job working together? hand, working alone, B would need 4
(a) 4 h
(b) 5 h
(c) 6 h
(d) 7 h
13. When a bucket is half full, the weight of the bucket and the water is 10 kg. When the bucket is twothirds full, the total weight is 11 kg. What is the total weight (in kg), when the bucket is completely full?
60° r
The smaller circle is tangent to the two bounding radii and the arc of the sector. The radius of the small circle is r 2 2 3r (c) 5 (a)
r 3 r (d) 2
(a) 12
(b) 12
1 2
(c) 12
2 3
(d) 13
14. How many ordered pairs of (m, n ) integers satisfy m 12 ? = 12 n
(b)
(a) 30
9. In the figure, AHKF , FKDE and HBCK are unit squares, AD and BF intersect in X. Then, the ratio of the areas of triangles AXF and ABF is A
E
F
(b) 15
(c) 12
(d) 10
15. Let S = {1, 2, 3, ..., 40) and let A be a subset of S such that no two elements in A have their sum divisible by 5. What is the maximum number of elements possible in A? (a) 10
(b) 13
(c) 17
(d) 20
X
PHYSICS H
16. A clay ball of mass m and speed v strikes another metal ball of same mass m, which is at rest. They stick together after collision. The kinetic energy of the system after collision is
C
B
1 (a) 4
D
K
1 (b) 5
(c)
1 6
(d)
1 8
10. Suppose Q is a point on the circle with centre P and radius 1, as shown in the figure, R is a point outside the circle such that QR = 1 and ∠QRP = 2°. Let S be the point where the segment RP intersects the given circle. Then, measure of ∠RQS equals
1 S
R
(b) 87°
(c) 88°
(d) 89°
hands of a clock make two angles between them whose sum is 360°. At 6:15 the difference between these two angles is (b) 170°
(c) 175°
(d) mv2
17. A ball falls vertically downward and bounces off a horizontal floor.The speed of the ball just before reaching the floor (u1 ) is equal to the speed just after leaving contact with the floor (u2 ), u1 = u2. The corresponding magnitudes of accelerations are denoted respectively by a1 and a 2. The air resistance during motion is proportional to speed and is not negligible. If g is acceleration due to gravity, then (b) a1 > a2 (d) a1 = a2 = g
flow of electrons in an electric circuit?
11. Observe that, at any instant, the minute and hour
(a) 165°
(c) 2 mv2
18. Which of the following statements is true about the
P
2°
(a) 86°
(b) mv2 / 4
(a) a1 < a2 (c) a1 = a2 ≠ g
Q 1
(a) mv2 / 2
(d) 180°
12. Two workers A and B are engaged to do a piece of work. Working alone, A takes 8 h more to complete the work than, if both worked together. On the other
(a) Electrons always flow from lower to higher potential (b) Electrons always flow from higher to lower potential (c) Electrons flow from lower to higher potential, except through power sources (d) Electrons flow from higher to lower potential, except through power sources
19. A boat crossing a river moves with a velocity v relative to still water. The river is flowing with a velocity v / 2 with respect to the bank. The angle with respect to the flow direction with which the boat should move to minimize the drift is (a) 30°
(b) 60°
(c) 150°
(d) 120°
110
KVPY Question Paper 2012 Stream : SA
20. In the Arctic region, hemispherical houses called Igloos are made of ice. It is possible to maintain a temperature inside an Igloo as high as 20°C because (a) ice has high thermal conductivity (b) ice has low thermal conductivity (c) ice has high specific heat (d) ice has higher density than water
S R
(a) P < Q < R < S (c)R < Q < P < S
21. In the figure below, PQRS denotes the path followed by a ray of light as it travels through three media in succession. The absolute refractive indices of the media are µ1 , µ 2 and µ3 , respectively. (The line segment RS in the figure is parallel to PQ). P
25. A circular metallic ring of radius R has a small gap of width d. The coefficient of thermal expansion of the metal is α in appropriate units. If we increase the temperature of the ring by an amount ∆T, then width of the gap
µ1 Q
(a) will increase by an amount dα∆T (b) will not change (c) will increase by an amount (2πR − d ) α∆T (d) will decrease by an amount dα∆T
µ2 R
µ3 S
26. A girl holds a book of mass m against a vertical wall with a horizontal force F using her finger, so that the book does not move. The frictional force on the book by the wall is
Then, (a) µ 1 > µ 2 > µ 3 (c µ 1 < µ 2 < µ 3
(b) P < Q < R = S (d) P > R > Q > S
(b) µ 1 = µ 3 < µ 2 (d) µ 1 < µ 3 < µ 2
(a) F and along the finger but pointing towards the girl (b) µF upwards, where µ is the coefficient of static friction (c) mg and upwards (d) equal and opposite to the resultant of F and mg
22. A ray of white light is incident on a spherical water drop whose centre is C as shown below.
27. A solid cube and a solid sphere both made of same C
material are completely submerged in water but to different depths. The sphere and the cube have same surface area. The buoyant force is
When observed from the opposite side, the emergent light (a) will be white and will emerge without deviating (b) will be internally reflected (c) will split into different colours such that the angles of deviation will be different for different colours (d) will split into different colours such that the angles of deviation will be the same for all colours
23. A convex lens of focal length 15 cm is placed in front of a plane mirror at a distance 25 cm from the mirror. Where on the optical axis and from the centre of the lens should a small object be placed such that the final image coincides with the object? (a) 15 cm and on the opposite side of the mirror (b) 15 cm between the mirror and the lens (c) 7.5 cm and on the opposite side of the mirror (d) 7.5 cm and between the mirror and the lens
24. Following figures show different combinations of identical bulb(s) connected to identical battery(ies). Which option is correct regarding the total power dissipated in the circuit?
(a) greater for the cube than the sphere (b) greater for the sphere than the cube (c) same for the sphere and the cube (d) greater for the object that is submerged deeper
28.
238 92 U
atom disintegrates to 214 84 Po with a half of . × 109 years by emitting six αparticles and 45 n electrons. Here, n is (a) 6
(b) 4
(c) 10
(d) 7
29. Which statement about the Rutherford model of the atom is not true? (a) There is a positively charged centre in an atom called the nucleus (b) Nearly all the mass of an atom resides in the nucleus (c) Size of the nucleus is comparable to the atom (d) Electrons occupy the space surrounding the nucleus
30. A girl brings a positively charged rod near a thin neutral stream of water from a tap. She observes that the water stream bends towards her. Instead, if she were to bring a negatively charged rod near to the stream, it will (a) bend in the same direction (b) bend in the opposite direction (c) not bend at all (d) bend in the opposite direction above and below the rod
P Q
111
KVPY Question Paper 2012 Stream : SA 39. The major product in the following reaction is
CHEMISTRY
H3 C— C ≡≡ C— H + HBr (excess)
31. The weight of calcium oxide formed by burning 20 g (a) 36 g
(b) 56 g
(c) 28 g
(a)
(d) 72 g
H3 C
32. The major products in the reaction Br3 CCHO → NaOH
O
(c) H3C
H
ONa
H
Br
(b) NaBr +
C
H
Br OH +
Br
ONa
Br O
40 + 33. The number of electrons plus neutrons in 19 K is
(b) 59
(c) 39
(d) 40
34. Among the following, the most basic oxide is (b) P2O5
(d) Na 2O
(c) SiO2
35. By dissolving 0.35 mole of sodium chloride in water, 1.30 L of salt solution is obtained. The molarity of the resulting solution should be reported as (a) 0.3 M (c) 0.27 M
Br
Br
(d) H3C
CH2
(b) 0.269 M (d) 0.2692 M
Br
constant KC , will proceed in the direction of the products when (a) QC = KC (c) QC > KC
(b) QC < KC (d) QC = 0
42. Acetyl salicylic acid is a pain killer and is commonly known as (a) paracetamol (c) ibuprofen
(b) aspirin (d) penicillin
43. The molecule which does not exhibit strong hydrogen bonding is (a) methyl amine (c) diethyl ether
(b) acetic acid (d) glucose
44. The following two compounds are
36. Among the quantities, density (ρ), temperature (T ),
H CH3
enthalpy (H ), heat capacity (C p ), volume (V ) and pressure ( p), a set of intensive variables are (a) ( ρ, T , H ) (c) (V , T , C p )
(b) (H , T , V ) (d) ( ρ, T , p )
37. The value of x in KAl(SO4 )x ⋅ 12H2O is (a) 1
(b) 2
(c) 3
(d) 4
38. Among the following substituted pyridines, the most basic compound is
H 3C
H3C
H
(a) geometrical isomers (b) positional isomers (c) functional group isomers (d) optical isomers
45. The graph that does not represent the behaviour of an ideal gas is
Me
Me
at constant p
at constant T
N
(a) (a)
CH
41. A reaction with reaction quotient QC and equilibrium
Br
Br
(a) Al 2O3
CH2
(a) CH3CONHCH2CH3 (b) CH3CH == NCH2CH3 (c) NH+3 CH2CH3 ⋅ CH3COO– (d) CH3CON == CHCH3
Br
(c) NaOBr +
(a) 38
Br
H C
CH 3CH 2 NH 2
O
(d)
CH3
is CH3 COOH →
H
Br
C
40. The major product in the following reaction at 25°C
C
D
(b) H3C
Br
Br
are (a) CHBr3+
Br
CH2
of calcium in excess oxygen is
(b) p
p
(b) N
N
CH3
Cl
(c) pV
(d)
(d) N
N
1/V at constant p
V
T
– 273°C
V
T
112
KVPY Question Paper 2012 Stream : SA 53. Myeloid tissue is a type of
BIOLOGY 46. A smear of blood from a healthy individual is stained with a nuclear stain called hematoxylin and then observed under a light microscope. Which of the following cell types would be highest in number? (a) Neutrophils (c) Eosinophils
(b) Lymphocytes (d) Monocytes
47. Which of the following biological phenomena involves a bacteriophage? (a) Transformation (c) Translocation
(b) Conjugation (d) Transduction
48. In which compartment of a cell does the process of
(a) haematopoietic tissue (c) muscular tissue
(b) cartilage tissue (d) areolar tissue
54. The heart of an amphibian is usually (a) twochambered (c) fourchambered
(b) threechambered (d) three and halfchambered
55. Gigantism and acromegaly are due to defects in the function of which of the following glands? (a) Adrenals (b) Thyroid
56. The pH of 10 (a) 8
−8
(c) Pancreas (d) Pituitary
M HCl solution is
(b) close to 7 (c) 1
glycolysis take place?
some of its own proteins?
(a) Golgi complex (c) Mitochondria
(a) Lysosome (c) Vacuole
(b) Cytoplasm (d) Ribosomes
49. Huntington’s disease is a disease of the (a) nervous system (c) respiratory system
(b) circulatory system (d) excretory system
50. A cell will experience the highest level of endosmosis when it is kept in (a) distilled water (c) salt solution
(b) sugar solution (d) protein solution
51. When the leaf of the ‘touchmenot’ (chuimui, Mimosa pudica) plant is touched, the leaf droops because (a) a nerve signal passes through the plant (b) the temperature of the plant increases (c) water is lost from the cells at the base of the leaf (d) the plant dies
52. If you are seeing mangroves around you, which part of India are you visiting? (a) Western Ghats (c) Sunderbans
(b) Thar desert (d) Himalayas
(d) 0
57. Which one of the following organelles can synthesise (b) Golgi apparatus (d) Mitochondrion
58. Maltose is a polymer of (a) one glucose and one fructose molecule (b) one glucose and one galactose molecule (c) two glucose molecules (d) two fructose molecules
59. The roots of some higher plants get associated with a fungal partner. The roots provide food to the fungus while the fungus supplies water to the roots. The structure so formed is known as (a) lichen (c) mycorrhiza
(b) Anabaena (d) Rhizobium
60. Prehistoric forms of life are found in fossils. The probability of finding fossils of more complex organisms (a) increases from lower to upper strata (b) decreases from lower to upper strata (c) remains constant in each stratum (d) uncertain
PARTII (2 Marks Questions) MATHEMATICS
64. A train leaves Pune at 7:30 am and reaches Mumbai
a 2+b is a b 2+c rational number, then which of the following is always an integer?
61. Let a , b, c be positive integers such that
(a)
2a 2 + b2 2b2 + c2
a 2 + b2 − c2 a 2 + 2b2 a 2 + b2 + c2 (b) (d) (c) 2 2 a+ b− c a+ b− c b + 2c
62. The number of solutions (x, y, z) to the system of equations x + 2 y + 4z = 9, 4 yz + 2xz + xy = 13, xyz = 3, such that at least two of x, y, z are integers is (a) 3
(b) 5
(c) 6
(d) 4
at 11:30 am. Another train leaves Mumbai at 9:30 am and reaches Pune at 1:00 pm. Assuming that the two trains travel at constant speeds, at what time do the two trains cross each other? (a) 10:20 am (b) 11:30 am (c) 10:26 am (d) data not sufficient
65. In the given figures, which has the shortest path? (a)
(b)
(c)
(d)
63. In a ∆ ABC, it is known that AB = AC. Suppose D is the midpoint of AC and BD = BC = 2. Then, the area of the ∆ ABC is (a) 2
(b) 2 2
(c) 7
(d) 2 7
113
KVPY Question Paper 2012 Stream : SA N
PHYSICS 66. In the circuit shown, nidentical resistors R are
connected in parallel (n > 1) and the combination is connected in series to another resistor R0. In the adjoining circuit n resistors of resistance R are all connected in series alongwith R0. R R R0
R
R
n R0
E
S
(a) N
(b) E
The batteries in both circuits are identical and net power dissipated in the n resistors in both circuits is same. The ratio R0 /R is (c) n 2
(d) 1/ n
67. A firecracker is thrown with velocity of 30 ms −1 in a direction which makes an angle of 75° with the vertical axis. At some point on its trajectory, the firecracker splits into two identical pieces in such a way that one piece falls 27 m far from the shooting point. Assuming that all trajectories are contained in the same plane, how far will the other piece fall from the shooting point? (Take, g = 10 ms −2 and neglect air resistance) (a) 63 m or 144 m (c) 28 m or 72 m
(c) W
(d) S
temperature 80°C. The latent heat of ice is 80 cal/g and the specific heat of water is 1 cal/g°C. Assuming no heat loss to the environment, the amount of ice which does not melt is (a) 100 g
E
(b) n
E
70. 150 g of ice is mixed with 100 g of water at
R R
(a) 1
W
A clockwise current is induced in the loop when loop is pulled towards
n R
i
(b) 0
71. Upon fully dissolving 2.0 g of a metal in sulphuric acid, 6.8 g of the metal sulphate is formed. The equivalent weight of the metal is (a) 13.6 g
(b) 20.0 g
(c) 4.0 g
0.1 M HCl and 0.2 M H2SO4, the concentration of H+ in the resulting solution is (a) 0.30 mol/L (b) 0.25 mol/L (c) 0.15 mol/L (d) 0.10 mol/L
73. The products X and Y in the following reaction sequence are NO2 Sn/HCl
X
(i) NaNO2/HCl
NO (a)
Y
Y:
NH2
Br
(b) X :
h
(ii) CuBr, ∆
Cl
X:
m
Y:
NO2
Br
θ
(b) µmgh / sin θ (d) µmgh / cosθ
(c)
X:
Y: Cl
69. A circular loop of wire is in the same plane as an infinitely long wire carrying a constant current i. Four possible motions of the loop are marked by N, E, W, and S as shown below.
(d) 10.0 g
72. Upon mixing equal volumes of aqueous solutions of
68. A block of mass m is sliding down an inclined plane
(a) µmgh (c) mgh
(d) 50 g
CHEMISTRY
(b) 72 m or 99 m (d) 63 m or 117 m
with constant speed.At a certain instant t0, its height above the ground is h. The coefficient of kinetic friction between the block and the plane is µ. If the block reaches the ground at a later instant t g , then the energy dissipated by friction in the time interval (t g − t0 ) is
(c) 150 g
Cl
NO
Cl
(d) X :
Y: Cl
Cl
114
KVPY Question Paper 2012 Stream : SA
74. A plot of the kinetic energy mv2 of ejected 1 2
BIOLOGY
electrons as a function of the frequency (ν ) of incident radiation for four alkali metals (M1, M 2, M3 , M 4 ) is shown below. M1 M2
M3
76. A baby is born with the normal number and distribution of rods, but no cones in his eyes. We would expect that the baby would be (a) colourblind (c) blind with both eyes
M4
(b) nightblind (d) blind with one eye
77. In mammals, pleural membranes cover the lungs as Kinetic energy
well as insides of the ribcage. The pleural fluid in between the two membranes (a) dissolves oxygen for transfer to the alveoli (b) dissolves CO2 for transfer to the blood (c) provides partial pressure (d) reduces the friction between the ribs and the lungs
v
The alkali metals M1, M 2, M3 and M 4 are, respectively
78. At which phase of the cell cycle, DNA polymerase activity is at its highest?
(a) Li, Na, K and Rb (b) Rb, K, Na and Li (c) Na, K, Li and Rb (d) Rb, Li, Na and K
(a) Gap 1 (G1 ) (c) Synthetic (S)
79. Usain Bolt, an olympic runner, at the end of a 100 metre sprint, will have more of which of the following in his muscles?
75. The number of moles of Br2 produced when two moles of potassium permanganate are treated with excess potassium bromide in aqueous acid medium is (a) 1 (b) 3 (c) 2 (d) 4
(c) (a) (b) (c) (b) (c)
(a) ATP (c) Lactic acid
(b) Pyruvic acid (d) Carbon dioxide
80. Desert temperature often varies between 050°C. The DNA polymerase isolated from a camel living in the desert will be able to synthesise DNA most efficiently at (a) 0°C
(b) 37°C
(c) 50°C
(d) 25°C
Answers
PARTI 1 11 21 31 41 51
(b) Mitotic (M) (d) Gap 2 (G 2)
2 12 22 32 42 52
(b) (c) (a) (a) (b) (c)
3 13 23 33 43 53
(b) (d) (a) (c) (c) (a)
4 14 24 34 44 54
(a) (a) (d) (d) (b) (b)
5 15 25 35 45 55
(d) (c) (a) (c) (b, c) (d)
6 16 26 36 46 56
(b) (b) (b) (d) (a) (b)
7 17 27 37 47 57
(b) (a) (b) (b) (d) (d)
8 18 28 38 48 58
(b) (c) (b) (b) (b) (c)
9 19 29 39 49 59
(b) (d) (c) (b) (a) (c)
10 20 30 40 50 60
(b) (b) (a) (c) (a) (a)
PARTII 61 71
(d) (b)
62 72
* No option is correct.
(b) (b)
63 73
(c) (b)
64 74
(c) (b)
65 75
(d) (*)
66 76
(a) (a)
67 77
(d) (d)
68 78
(c) (c)
69 79
(b) (c)
70 80
(d) (b)
115
KVPY Question Paper 2012 Stream : SA
Solutions 1. (c) Let f (x) = ax2 + bx + c
5. (d) We have,
[Q f (x) is quadratic polynomial] …(i) f (2) = 4a + 2b + c = 10 [Q f (2) = 10] …(ii) f (−2) = 4a − 2b + c = − 2 [Q f (−2) = − 2] On subtracting Eq. (ii) from Eq.(i), we get 4b = 12 ⇒ b = 3 ∴Coefficient of x in f (x) = b = 3
2. (b) Let x = 0.75 According to the question, x3 + (x + x2 + 1) 1− x =
x3 + (1 − x)(1 + x + x2 ) 1− x
=
x3 + 1 − x3 1 = 1− x 1− x
Now, put the value of x 1 1 = 1 − 0.75 0.25 100 = =4 25 So, square root of the equation =
22 + 42 + 62 + ... + (2n )2 12 + 32 + 52 + ... + (2n − 1)2
⇒ ⇒
3. (b) Given, sides of triangle are positive integer in an AP and the smallest side is 10. ∴Sides of triangle are 10, 10 + d , 10 + 2d , d ∈N We know in triangle sum of two sides is greater than third sides. …(i) ∴ 10 + 10 + d > 10 + 2d …(ii) 10 + 10 + 2d > 10 + d …(iii) 10 + d + 10 + 2d > 10 From Eqs. (i), (ii) and (iii), we get d < 10 ∴ d = 1, 2, 3, 4, 5, 6, 7, 8, 9 Hence, there are 9 triangles possible.
Σ(2n − 1)2 4Σ n 2 Σ(4n 2 − 4n + 1)
>
>
101 100
101 100
4Σ n 2
101 > 4Σn − 4 Σn + Σ1 100 4(n )(n + 1)(2n + 1) ⇒ 6 101 > 4n (n + 1)(2n + 1) 4n (n + 1) 100 − +n 6 2 4n (n + 1)(2n + 1) 101 ⇒ > n [4(2n 2 + 3n + 1) − 12n − 12 + 6] 100
⇒
⇒
4=2
Σ(2n )2
⇒ ⇒
> 101 .
2
4(n + 1)(2n + 1) 101 > 100 8n 2 − 2 4(2n + 1)(n + 1) 101 > 2(2n + 1)(2n − 1) 100 2n + 2 101 > 2n − 1 100
7. (b) Given, D
⇒ ⇒
200n + 200 > 202n − 101 2n < 301 301 n< ⇒ 2 ∴Maximum value of n = 150
x
6. (b) Given,
a = 3k, b = k, c = k, d = k a 3k ∴ = b + 2c + 3d k + 2k + 3k =
3k 1 = 6k 2
x
U
1–2x
T
x
C
45° 1–2x
V
S
W
R
A
A
F
E
I
B
D
C
4. (a) We have, a a+b a+ b+ c a+b+c+d = = = =k 3 4 5 6 On solving, we get
⇒ ∠FBD + ∠FID = 180° [Q sum of interior opposite angle of cyclic quadrilateral is 180°] Now, ∠FID = ∠ADC + ∠ICD [Q sum angle properties] ⇒ ∠FID = ∠BAD + ∠ABD + ∠ICD A C = + B+ 2 2 A + C + 2B = 2 A + B +C + B = 2 180° + B = 2 ∴∠FBD + ∠FID = 180° 180° + B =B+ = 180° 2 ⇒ 3B = 180° ⇒ B = 60° ∠B 60 = = 30° ∴∠IFD = 2 2
AD , BE , CF are angle bisectors of angle A , B and C respectively. I is the concurrent point of angle bisector and BD IF are concyclic. Now, BDIF is concyclic. ∠B ∠IFD = ∠IBD = ∴ 2 [Q angle on same segments are equals ]
P
Q
B
ABCD is square of length 1 unit and a regular octagon is formed by cutting congruent isosceles triangle. Let DUV is isosceles right angled triangle. ∴ ∠D = 90° ∴ ∠DVU = 45° In ∆DVU, x cos45° = 1 − 2x 1 x = ⇒ 2 1 − 2x 1 − 2x 2= ⇒ x 1 ⇒ 2 = −2 x 1 = 2+ 2 ⇒ x
116 ⇒
KVPY Question Paper 2012 Stream : SA x=
2 −1 2
1 = 2 ( 2 + 1)
In ∆AXF and ∆BAF, (common) ∠F = ∠F ∠X = ∠A = (90° ) ∴ ∆AXF ~ ∆BAF ar(∆AXF ) AF 2 1 = = ∴ ar(BAF ) BF 2 5
∴ Side of regular octagon = 1 − 2x 2( 2 − 1) = 1− 2 = 1 − 2 ( 2 − 1) = 1−2 + 2 = 2−1
Let the complete work in t h 1 1 t + 8 t + = x y x t 8 = ⇒ y x
10. (b) Given,
8. (b) Given, OA = OB = r Q
B
1
1
2° S
R
P
C
R
R
Q
r
B
∠QRP = 2° PQ = QR = 1 ∠QPR = 2° ∠RQP = 180° − 4° = 176° SP = SQ radii of circle ∠SQP = ∠QSP
R
A
R
60° 30°
O
r
∴ A
P
Also, OC is radius of sector ∴ OC = r Now, OC = OB + BC [BC is radius of circle] OC = OB + R In ∆OPB, BP [Q ∠BOP = 30°] sin 30° = OB R 1 ⇒ = 2 OB ⇒ OB = 2R ∴ OC = 2R + R = 3R ⇒ r = 3R r R= ⇒ 3
∴
180° − 2° 2 178° = = 89° 2 ∠RQS = ∠RQP − ∠SQP = 176° − 89° = 87° =
11. (a) At 6: 15, 12
3
9
r
9. (b) We have, 1
A
F
1
1
1
K
1
1
B
C
1
E
6
X
1 H
12. (c) Let the time taken by A to complete the job = x h and time taken by B to complete the job = y h A and B together works then they 1 1 complete the work = + in 1 h x y
1
D
AHKF , FKDE and HBCK is a unit square. AD and BF intersect at X. In ∆ABF, AB = 2 AF =1 ∴ FB 2 = AB 2 + AF 2 = 4 + 1 = 5
the minutes hand makes an angle is α. 1 ° α = 90° + 15 × ∴ 2 195 ° α = 2 and hour hand is β. Given, α + β = 360° ∴ β = 360° − α 195 ° 525 ° = 360° − = 2 2 Difference between their angles 525 195 330 = − = = 165° 2 2 2
⇒
…(i)
1 [B would 4 h more to 2 complete the work] 9 t+ 1 1 2 t + = x y y t 9 …(ii) = x 2y
From Eqs. (i) and (ii), t 2 = 36 ⇒ t =6h
13. (d) Let the weight of bucket be x kg and the weight of water completely full be y kg. According to the problem, y …(i) x + = 10 2 2y and …(ii) x+ = 11 3 On solving Eqs. (i) and (ii), we get x = 7, y = 6 ∴Total weight, when bucket is completely full is (x + y) kg i.e. 7 + 6 = 13 kg 14. (a) We have, m 12 = 12 n ⇒ mn = 144 ⇒ mn = 24 × 32 Total number of divisor of 144 is (4 + 1)(2 + 1) = 15 When m and n are positive integers. If m and n are negative integers, then also number of divisor is 15. ∴Total ordered pairs of (m, n ) when m and n are integers = 15 + 15 = 30
15. (c) We have, S = {1, 2, 3, 4,..., 40} A is subset of S whose sum of two element of A is not divisible by 5. Possible set A = { 1, 2, 5, 6, 7, 11, 12, 16, 17, 21, 22, 26, 27, 31, 32, 36, 37} ∴ Maximum number of elements in A is 17.
117
KVPY Question Paper 2012 Stream : SA 19. (d) To minimize drift, let angle at
16. (b) Given situation is Initially one of u2=0 the ball is at rest
m
m
u 1 =v
Finally both balls moves with same speed together
v
2
17. (a) Air resistance is same in both case.When ball is moving down, air resistance is directed away from g.
θ
(Q)
90°
From above velocity triangle, v/2 1 sin θ = = or θ = 30° v 2 So, angle with respect to direction of flow is 90° + θ° = 90° + 30° = 120°.
20. (b) Ice is a bad conductor of heat, its
(R)
thermal conductivity is very low. So, no exchange of heat from outside surrounding occurs in an Igloo. Thermal conductivity of ice is 16 . Wm −1 K −1 .
g
So, acceleration of ball moving downwards is mg − kv a1 = m k or a1 = g − v m where, k is constant. When ball is moving up, air resistance and g both are directed downwards. v
µ1
Q
3V 2 V2 = R /3 R
P =
+ –
P = + –
V2 R
+ –
(S)
at point Q and it bends away from normal at point R. v
V2 3R
vres
21. (b) Light ray bends towards normal
Air resistance
P =
+ –
v
Conservation of momentum, gives v mv = 2 mV ⇒ V = 2 So, kinetic energy after collision is 1 1 v (2m) V 2 = × 2m × 2 2 2 1 = mv2 4
(P)
v/2
m m
Kf =
24. (d) Power consumed in each case is
which boat is directed by θ, as shown below.
+ –
P =
(2V ) 2 4V 2 = R R
– +
So, order of increasing power consumption is P > R > Q > S.
µ2
25. (a) Gap or cavity also expands at
µ3
R
Pa ra ra y
llel
s
same rate as that of metal. Hence, width of gap also increases by same amount. ∴Width of gap increases by ∆d = dα ∆T . f=µN
26. (b) Also, the emergent ray is parallel to incident ray. Hence, µ 1 = µ 3 < µ 2 is correct option.
F
22. (a) As, incident light is normal to the surface, so no deviation or dispersion occurs. Air resistance
g
So,acceleration while moving upwards is k v a2 = g + m Clearly, a2 > a1 .
18. (c) The free electrons experiences electrostatic force in the direction opposite to the direction of electric fired being is of negative charge. The electric field always directed from higher potential to lower potential. Therefore electrostatic force and negative charge or electrons always flows from lower to higher potential until the potantials become equal. Hence, option (c) is correct.
Emergent beam
N
Incident beam C
23. (a) If object is placed
mg
Minimum force F must be such that generated friction is able to balance weight mg of book. So, f = µN = µF.
27. (b) Given, Surface area of cube = Surface area of sphere O r
at the focus of lens on opposite side of mirror, then light rays after refraction from lens become parallel to the principal axis. These parallel rays are reflected back over same path and again converges at focus.
a
⇒ ⇒
6a 2 = 4 πr 2 4π a = = 6 r
…(i) 2π 3
118
KVPY Question Paper 2012 Stream : SA
Now, buoyant force is FB = V in .ρf . g So, ratio of buoyant force on cube and sphere is (FB )cube V = cube (FB )sphere V sphere 3
3 2π 2 = = × 4 3 3 4 π πr 3 a3
= ∴
9 × 8 × π3 16 × π 2 × 27
=
π 6
(FB )cube < (FB )sphere
28. (b) Decay is 4 − → 214 84 Po + 6 2 He + ne Conservation of mass number and atomic number gives, ⇒ 92 = 84 + 12 − n ⇒ n = 4 238 92 U
29. (c) In Rutherford model, positive charge acquires a very small place at centre of atom in nucleus. So, option (c) is incorrect.
30. (a) Neutral objects are always attracted towards both positively and negatively charged objects. So, water stream still bends in same direction. ∆
31. (c) 2Ca + O2 → 2CaO Excess
20 1 = mole 40 2 1 mole of Ca produces 1 mole of CaO 1 1 ∴ mole of Ca will produces mole 2 2 of CaO 1 mole of CaO = 56 g 1 1 mole of CaO = 56 × = 28 g 2 2 Thus, 28 g of CaO is formed by burning 20 g of Ca in excess oxygen. Number of moles of Ca =
32. (a) NaOH
Br3 CCHO → CHBr3 + HC O− Na +  O This reaction is known as bromoform reaction where the carbonyl carbon gets oxidised by sodium hydroxide to sodium salts of corresponding carboxylic acid having one carbon atom less than of carbonyl compound and the methyl group is converted to bromoform. 40 + 33. (c) For, 19 K
Number of electrons in K+ = 18 Number of neutrons in K+ = 40 − 19 = 21 ⇒ Sum of electron + neutron = 18 + 21 = 39
34. (d) All the central atom of the given oxide belong to 2nd period, as we move from left to right in a period the basicity of oxide decreases. ∴The order of increasing basicity of oxide would be P2O5 < SiO2 < Al 2O3 < Na 2O Thus, the most basic oxide would be Na 2O.
The increasing order of basicity of compounds given in options will be Me Cl
CH3
T1
(b) T3 > T2 > T1 (d) T3 > T2 = T1
21. We sit in the room with windows open. Then, (a) air pressure on the floor of the room equals the atmospheric pressure but the air pressure on the ceiling is negligible (b) air pressure is nearly the same on the floor, the walls and the ceiling (c) air pressure on the floor equals the weight of the air column inside the room (from floor to ceiling) per unit area (d) air pressure on the walls is zero, since the weight of air acts downward
22. A girl standing at point P on a beach wishes to reach a point Q in the sea as quickly as possible. She can run at 6 kmh −1 on the beach and swim at 4 kmh −1 in the sea. She should take the path Q
Sea
A
B
C
D
C
A
18. A juggler tosses a ball up in the air with initial speed
The minimum value of the refractive index of the prism is close to (a) 1.10
(b) 1.55
(c) 1.42
(d) 1.72
24. A convex lens is used to form an image of an object on a screen. If the upper half of the lens is blackened, so that it becomes opaque, then (a) only half of the image will be visible (b) the image position shifts towards the lens (c) the image position shifts away from the lens (d) the brightness of the image reduces
25. A cylindrical copper rod has length L and resistance R. If it is melted and formed into another rod of length 2L, then the resistance will be (a) R
(b) 2R
(c) 4R
(d) 8R
26. Two charges +Q and −2Q are located at points A and B on a horizontal line as shown below. +Q
–2Q
A
B
The electric field is zero at a point which is located at a finite distance (a) on the perpendicular bisector of AB (b) left of A on the line (c) between A and B on the line (d) right of B on the line
27. A 750 W motor drives a pump which lifts 300 L of water per minute to a height of 6 m. The efficiency of the motor is nearly (Take, acceleration due to gravity to be 10 m / s2) (a) 30%
(b) 40%
(c) 50%
(d) 20%
28. Figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion PQ is 3A P
Q
1A
2A
Beach
8A
P
(a) PAQ
(b) PBQ
(c) PCQ
(d) PDQ
4A
2A
23. Light enters an isosceles right triangular prism at normal incidence through face AB and undergoes total internal reflection at face BC as shown below.
(a) zero (c) 4 A from Q to P
(b) 3 A from P to Q (d) 6 A from Q to P
KVPY Question Paper 2011 Stream : SA
126 29. A nucleus of lead Pb214 82 emits two electrons followed by an αparticle. The resulting nucleus will have (a) 82 protons and 128 neutrons (b) 80 protons and 130 neutrons (c) 82 protons and 130 neutrons (d) 78 protons and 134 neutrons
(a) zero, positive (c) positive, zero
(c) 3 × 1027
(d) 6 × 1030
(NH4 )2 Cr2O7 is (a) oxygen (c) ammonia
CHEMISTRY 31. Two balloons A and B containing 0.2 mole and 0.1 mole of helium at room temperature and 2.0 atm, respectively, are connected. When equilibrium is established, the final pressure of He in the system is (a) 1.0 atm (c) 0.5 atm
(b) nitric oxide (d) nitrogen
37. The solubility curve of KNO3 in water is shown below. Solubility (gram)/100 gram of water
room at standard temperature and pressure is of the order of (b) 3 × 1024
(b) zero, negative (d) negative, zero
36. The gas produced from thermal decomposition of
30. The number of air molecules in a (5m × 5m × 4m) (a) 6 × 1023
When the partition is removed, the gases mix. The changes in enthalpy (∆H ) and entropy (∆S ) in the process, respectively, are
(b) 1.5 atm (d) 2.0 atm
32. In the following set of aromatic compounds NO2
250 200 150 100 50 0
0
10 20 30 40 Temperature (°C)
50
The amount of KNO3 that dissolves in 50 g of water at 40°C is closest to (i)
(ii)
COOCH3
OCH3
(a) 100 g
(b) 150 g
(a) 2pentanone (c) 3pentanol (iii)
(iv)
The correct order of reactivity toward FriedelCrafts alkylation is (a) i >ii > iii > iv (b) ii > iv > iii > i (c) iv > ii > iii > i (d) iii > i > iv > ii
33. The set of principal (n ), azimuthal (l) and magnetic (ml ) quantum numbers that is not allowed for the electron in Hatom is (a) n = 3, l = 1, ml = − 1 (b) n = 3, l = 0, ml = 0 (c) n = 2, l = 1, ml = 0 (d) n = 2, l = 2, ml = − 1
(c) 200 g
(d) 50 g
38. A compound that shows positive iodoform test is (b) 3pentanone (d) 1pentanol
39. After 2 hours the amount of a certain radioactive substance reduces to 1/16th of the original amount (the decay process follows firstorder kinetics). The halflife of the radioactive substance is (a) 15 min
(b) 30 min
(c) 45 min
(d) 60 min
40. In the conversion of a zinc ore to zinc metal, the process of roasting involves (a) ZnCO3 → ZnO (c) ZnS → ZnO
(b) ZnO → ZnSO4 (d) ZnS → ZnSO4
41. The number of P–H bond(s) in H3 PO2 , H3 PO3 and H3 PO4 , respectively, is (a) 2, 0, 1
34. At 298 K, assuming ideal behaviour, the average kinetic energy of a deuterium molecule is (a) two times that of a hydrogen molecule (b) four times that of a hydrogen molecule (c) half of that of a hydrogen molecule (d) same as that of a hydrogen molecule
35. An isolated box, equally partitioned contains two ideal gases A and B as shown
(b) 1, 1, 1
(c) 2, 0, 0
(d) 2, 1, 0
42. When chlorine gas is passed through an aqueous solution of KBr, the solution turns orange brown due to the formation of (a) KCl
(b) HCl
(c) HBr
(d) Br 2
43. Among H O
N
N
A 1 atm 25°C
B 1 atm 25°C
(i)
(ii)
(iii)
the compound which is not aromatic is (a) i (b) ii (c) iii (d) iv
(iv)
127
KVPY Question Paper 2011 Stream : SA 49. The gall bladder is involved in
44. Among the following compounds
(a) synthesising bile (b) storing and secreting bile (c) degrading bile (d) producing insulin
H 3C CH3
H3C
H 3C
CH3
H3C
CH3 (i)
(ii)
50. Which one of the following colours is the least useful
CH3
for plant life? (a) Red (c) Green
H3C CH3
CH3
51. At rest, the volume of air that moves in and out per
CH3 CH3
H3C
breath is called
H 3C
(a) resting volume (c) lung capacity
CH3 (iv)
(iii)
(b) vital capacity (d) tidal volume
52. How many sex chromosomes does a normal human inherit from father?
2,3dimethylhexane is (a) i (b) ii
(c) iii
(d) iv
(a) 1 (c) 23
45. The major product formed in the reaction,
(b) 2 (d) 46
53. In the 16th century, sailors who travelled long
Br Cl
NaCN DMSO, heat
distances had diseases related to malnutrition, because they were not able to eat fresh vegetables and fruits for months at a time. Scurvy is a result of the deficiency of
product
(a) carbohydrates (c) vitaminC
I
is
(b) Blue (d) Violet
Br
CN
(b) proteins (d) vitaminD
54. Which of the following structures is not found in plant cells?
Cl
(i)
Cl
(ii)
CN
CN Br
(b) Nucleus (d) Endoplasmic reticulum
55. The cell that transfers information about pain to the brain is called a
CN Cl
CN (iv)
(iii)
(a) Vacuole (c) Centriole
(a) neuron (c) histoblast
(b) blastocyst (d) haemocyte
56. The presence of nutrients in the food can be tested. (a) i
Benedict’s test is used to detect
I
I
(b) ii
(c) iii
(d) iv
(a) sucrose (c) fatty acid
(b) glucose (d) vitamins
57. Several minerals such as iron, iodine, calcium and
BIOLOGY 46. If parents have free earlobes and the offspring has attached earlobes, then the parents must be (a) homozygous (c) codominant
(b) heterozygous (d) nullizygous
47. During meiosis, there is (a) one round of DNA replication and one division (b) two rounds of DNA replication and one division (c) two rounds of DNA replication and two divisions (d) one round of DNA replication and two divisions
48. Blood clotting involves the conversion of (a) prothrombin to thromboplastin (b) thromboplastin to prothrombin (c) fibrinogen to fibrin (d) fibrin to fibrinogen
phosphorus are important nutrients. Iodine is found in (a) thyroxine (c) insulin
(b) adrenaline (d) testosterone
58. The principle upon which a lactometer works is (a) viscosity (c) surface tension
(b) density (d) presence of protein
59. Mammalian liver cells will swell up when kept in (a) hypertonic solution (c) isotonic solution
(b) hypotonic solution (d) isothermal solution
60. The form of cancer called ‘carcinoma’ is associated with (a) lymph cells (c) blood cells
(b) mesodermal cells (d) epithelial cells
KVPY Question Paper 2011 Stream : SA
128
PARTII (2 Marks Questions) MATHEMATICS 61. Let f (x) = ax + bx + c, where a , b, c are integers, 2
Suppose f (1) = 0, 40 < f (6) < 50, 60 < f (7) < 70 and 1000t < f (50) < 1000 (t + 1) for some integer t. Then, the value of t is (a) 2
(b) 3
(c) 4
(d) 5 or more
62. The expression 22 + 1 2 −1 2
+
32 + 1 3 −1 2
+
42 + 1 4 −1 2
+ .... +
(2011)2 + 1 (2011)2 − 1
ρ1 x1 −ρ2x2 x1 − x2 ρ1 x2 + ρ2x1 (c) ρ = x1 + x2
ρ1 x2 −ρ2x1 x2 − x1 ρ1 x1 + ρ2x2 (d) ρ = x1 + x2
(a) ρ =
(b) ρ =
67. A body of 0.5 kg moves along the positive Xaxis
lies in the interval 1 (a) 2010, 2010 2 1 (c) 2011,2011 2
When the spring object system is immersed in a liquid of density ρ1 as shown in the above figure, the spring stretches by an amount x1 (ρ > ρ1 ). When the experiment is repeated in a liquid of density (ρ2 < ρ1 ), the spring stretches by an amount x2. Neglecting any buoyant force on the spring, the density of the object is
1 1 (b) 2011 − , 2011 − 2012 2011 1 (d) 2012, 2012 2
under the influence of a varying force F ( in newton) as shown below. 3
63. The diameter of one of the bases of a truncated cone
(a) 65
(b) 55
(c) 45
F(N)
is 100 mm. If the diameter of this base is increased by 21% such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by 21%. The radius of the other base (in mm) is
2 1
(d) 35
0,0
2
4
simultaneously on motorcycles to meet each other. The speed of A is 3 times that of B. The distance between them decreases at the rate of 2 km per minute. Ten minutes after they start, A’s vehicle breaks down and A stops and waits for B to arrive. After how much time (in minutes) A started riding, does B meet A? (a) 15
(b) 20
(c) 25
(d) 30
65. Three taps A, B, C fill up a tank independently in 10 h, 20 h, 30 h, respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank? (a) 8
(b) 9
(c) 10
(d) 11
PHYSICS
6
8
10
x(m)
64. Two friends A and B are 30 km apart and they start
If the speed of the object at x =4 m is 316 . ms−1, then its speed at x = 8 m is
(a) 3.16 ms−1 (b) 9.3 ms−1
(c) 8 ms−1
(d) 6.8 ms−1
68. In a thermally isolated system, two boxes filled with an ideal gas are connected by a valve. When the valve is in closed position, states of the box 1 and 2 respectively, are (1 atm, V, T) and (0.5 atm, 4V, T). When the valve is opened, then the final pressure of the system is approximately (a) 0.5 atm
(b) 0.6 atm
(c) 0.75 atm (d) 1.0 atm
69. A student sees the top edge and the bottom centre C of a pool simultaneously from an angle θ above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the 4 h 7 pool is . If = , then cos θ is 3 x 4
66. An object with uniform density ρ is attached to a spring that is known to stretch linearly with applied force as shown below.
θ
h
C x
2 (a) 7
8 (b) 3 45
(c)
8 3 53
(d)
8 21
129
KVPY Question Paper 2011 Stream : SA 70. In the following circuit, 1Ω resistor dissipates power P . If the resistor is replaced by 9 Ω, the power dissipated in it is 3Ω
10 V
(a) P
+ –
1Ω
(b) 3P
(c) 9P
(d)
P 3
CHEMISTRY 71. An aqueous buffer is prepared by adding 100 mL of
0.1 mol L−1 acetic acid to 50 mL of 0.2 mol L−1 of sodium acetate. If pK a of acetic acid is 4.76, the pH of the buffer is
(a) 4.26
(b) 5.76
(c) 3.76
(d) 4.76
72. The maximum number of structural isomers possible for the hydrocarbon having the molecular formula C4H6, is (a) 12
(b) 3
(c) 9
(d) 5
73. In the following reaction sequence, X and Y ,
BIOLOGY 76. You remove four fresh tobacco leaves of similar size and age. Leave ‘leaf 1’ as it is, smear ‘leaf 2’ with vaseline on the upper surface, ‘leaf 3’ on the lower surface and ‘leaf 4’ on both the surfaces. Hang the leaves for a few hours and you observe that ‘leaf 1’ wilts the most, ‘leaf 2’ has wilted, ‘leaf 3’ wilted less than ‘leaf 2’ and ‘leaf 4’ remains fresh. Which of the following conclusions is most logical? (a) Tobacco leaf has more stomata on the upper surface (b) Tobacco leaf has more stomata on the lower surface (c) Stomata are equally distributed in upper and lower surfaces (d) No conclusion on stomatal distribution can be drawn from this experiment
77. Vestigial organs such as the appendix exist because (a) they had an important function during development which is not needed in the adult (b) they have a redundant role to play if an organ with similar function fails (c) nature cannot get rid of structures that have already formed (d) they were inherited from an evolutionary ancestor in which they were functional
78. Mendel showed that unit factors, now called alleles,
respectively, are O X
Y
OH
(b) C6 H5 COOH; LiAlH4 (a) H2O2 ; LiAlH4 (c) C6 H5 COOH;Zn /Hg ⋅ HCl (d) alk. KMnO4 ; LiAlH4
74. Among (i) [Co(NH3 )6 ]Cl3 , (ii) [Ni(NH3 )6 ]Cl2,
exhibit a dominant/recessive relationship. In a monohybrid cross, the .......... trait disappears in the first filial generation. (a) dominant (c) recessive
(b) codominant (d) semidominant
79. If a man with an Xlinked dominant disease has six
(iii) [Cr(H2O)6 ]Cl3 , (iv) [Fe(H2O)6 ]Cl2 the complex which is diamagnetic is
sons with a woman having a normal complement of genes, then the sons will
(a) i
(a) not show any symptoms of the disease (b) show strong symtpoms of the disease (c) three will show a disease symptom, while three will not (d) five will show a disease symptom, while one will not
(b) ii
(c) iii
(d) iv
75. At 783 K in the reaction, H2 ( g) + I2 ( g)
2HI( g), the molar concentrations (mol L−1 ) of H2 , I2 and HI at some instant of time are 0.1, 0.2 and 0.4, respectively. If the equilibrium constant is 46 at the same temperature, then as the reaction proceeds
e
(a) the amount of HI will increase (b) the amount of HI will decrease (c) the amount of H 2 and I2 will increase (d) the amount of H 2 and I2 will not change
80. In evolutionary terms, an Indian school boy is more closely related to (a) an Indian frog (b) an American snake (c) a Chinese horse (d) an African shark
KVPY Question Paper 2011 Stream : SA
130
Answers PARTI 1
(a)
2
(c)
3
(a)
4
(b)
5
(b)
6
(d)
7
(a)
8
(a)
9
(a)
10
(c)
11
(b)
12
(c)
13
(b)
14
(a)
15
(c)
16
(a)
17
(a)
18
(c)
19
(d)
20
(a)
21
(b)
22
(c)
23
(c)
24
(d)
25
(c)
26
(b)
27
(b)
28
(d)
29
(a)
30
(c)
31
(d)
32
(c)
33
(d)
34
(d)
35
(a)
36
(d)
37
(a)
38
(a)
39
(b)
40
(c)
41
(d)
42
(d)
43
(b)
44
(b)
45
(c)
46
(b)
47
(d)
48
(c)
49
(b)
50
(c)
51
(d)
52
(a)
53
(c)
54
(c)
55
(a)
56
(b)
57
(a)
58
(b)
59
(b)
60
(d)
PARTII 61
(c)
62
(c)
63
(b)
64
(d)
65
(a)
66
(b)
67
(d)
68
(b)
69
(c)
70
(a)
71
(d)
72
(c)
73
(b)
74
(a)
75
(a)
76
(b)
77
(d)
78
(c)
79
(a)
80
(c)
Solutions 1. (a) Given, P (x ) =
(x − b)(x − c) (x − c)(x − a ) + (a − b)(a − c) (b − c)(b − a ) (x − a )(x − b) + (c − a )(c − b) P (a ) = 1 + 0 + 0 = 1
Now, squaring both sides, we get 2 x + 1 = a 2 x 1 2 x + 2 + 2 = a2 ⇒ x On cubing both sides, we get x +
P (b) = 0 + 1 + 0 = 1 P (c) = 0 + 0 + 1 = 1 P (x) is a polynomial of degree atmost 2 and also attains same value i.e. 1 for distinct values of x (i.e. a , b, c). ∴ P (x) is an identity with only value equal to 1 for all R. (x − b)(x − c) (x − c)(x − a ) ∴ + + (a − b)(a − c) (b − c)(b − a ) (x − a )(x − b) =1 (c − a )(c − b)
2. (c) Given, a 2 + b2 = 81 ⇒
x2 + y2 = 121
⇒
ax + by = 99
Now, (a 2 + b2 )(x2 + y2 ) = 81 × 121 a x + b y + a y + b x = 81 × 121 2 2
and
2 2
2 2
2 2
…(i)
⇒
(ax + by) = 99
⇒
a 2x2 + b2 y2 + 2axby = 992
2
a 2 y2 + b2x2 − 2axby = 0 ⇒ ⇒
5. (b) Given,
0< r < 4 x
On subtracting Eq. (i) from Eq. (ii), we get (ay − bx)2 = 0
ay − bx = 0 1 1 3. (a) Given, x + = a and x2 + 3 = b x x
6. (d) Given, ABC is right angled triangle with B is 90°. A
6
4. (b) Given,a − b = 2,b − c = 3 and c − d = 4 ∴ a − b = ± 2, b − c = ± 3 and c − d = ± 4 Possible value of (a − d ) are ±9, ± 5, ± 3, ± 1 . ∴ a − d = 9, 5, 3, 1 Sum ofa − d = 9 + 5 + 3 + 1 = 18
⇒
D
B
C
AD is angle bisector of ∠ A. ∴
AB BD = AC DC
⇒
AB ⋅ CD = BD ⋅ AC
Area of ∆ADC = 10 1 × AB ⋅ CD = 10 2
x
⇒
1 × BD ⋅ AC = 10 2
x
⇒
1+ r = 9 5 π r 9 B = 1+ = 17 5 9 x C = (1 + 2r ) = 5 D = 1 +
90°
⇒
r A = 5 1+ = 9 π …(ii)
1 3 = a …(ii) x
On adding Eqs. (i) and (ii), we get x2 + 1 + x3 + 1 + 2 + 3 x + 1 x x3 x2 = a3 + a 2 1 3 3 ⇒ b + x + 2 + 2 + 3a = a + a 2 x 1 3 ⇒ x + 2 = a3 + a 2 − 3a − b − 2 x
(ax + by) = 99 2
3
1 1 3 = x + 3 + 3 x + x x
…(i)
All A , B ,C , D are in the form of (a )x = b x is largest when a is smallest. ∴ In A , B , C , D 0< r < 4 r is smallest 1+ 17 ∴Option (b) is correct.
x
1 9 = 5 r
BD =
⇒
BD =
20 6
⇒
BD =
10 3
20 AC [Q AC = 6]
131
KVPY Question Paper 2011 Stream : SA 7. (a) We know, θ =
AB OB
10. (c) We have, (512)3 − (253)3 − (259)3 ⇒
O θ
l=13
12
Now,
(512)3 + (−253)3 + (−259)3
We know that, a + b + c = 0 then a + b + c = 3abc
2 π (5) = 13θ ⇒θ =
= 3(512)(−253)(−259) = 3 ⋅ 512 ⋅ 253 ⋅ 259 = 3 ⋅ 29 ⋅ 11 × 23 × 7 × 37
10 π 13
11. (b) Given, square base pyramid is incomplete.
A 90°–θ
θ
B
x
90°–θ E(20–x) 20
8 D
AB ED = BE CD 12 20 − x ⇒ = x 8 ⇒ 96 = 20x − x2 ⇒ x2 − 20x + 96 = 0 ⇒ (x − 12)(x − 8) = 0 ⇒ x = 8, 12 Hence, x has two values 8 and 12 and their difference is 4.
∴
9. (a) Given, radius of each circle = 1
O 1 d 1 A
60° 1
The top layer = 13 balls There are 18 layer completed. So, total number of balls N = 132 + 142 + 152 + 162 + ... + 302 N = (12 + 22 + 32 + 42 + ... + 302 ) − (12 + 22 + 32 ... 122 ) 30 × 31 × 61 12 × 13 × 25 − ⇒ N = 6 6 ⇒ N = 5 × 31 × 61 − 2 × 13 × 25 = 9455 − 650 = 8805 ∴ 8000 < N < 9000
12. (c) Let the total distance = x x 6 x x x Water distance = − = 2 6 3 x Tar distance = 2 Speed travelling in mud = 3y Muddy distance =
1 1
1
D 1
1st
A(4, 3) 5
O (0, 0)
B
1
touch externally ∴ OA = OB = AB = 2 OD In ∆OAD, sin 60° = OA 3 = 3 ⇒ OD = OA sin 60° = 2 × 2 ∴ d = 1 + OD + 1= 1 + 3 + 1 = 3 + 2
After 1st jump position of frog at (4, 3). At 2nd jump position of frog at (0,6). At 3rd jump position of frog at (0,1). ∴Minimum number of jumps required for the frog to go from (0,0) to (0,1) and each distance is 5 units is 3.
14. (a) Digit 1 appears in 1, 10, 11, and
C
θ
(0, 1)
∴ There are 6 distinct prime divisors.
8. (a) ∆ABE ~ ∆EDC
12
2nd
5
3
∴ (512) − (253)3 − (259)3
5
From first figure, AB = 2 πr OB = 13 ∴
3
3
B
(0, 6)
512 − 253 − 259 = 0 3
A
13. (b) We have, initial position of frog = (0,0)
Speed travelling by stream = 4y Speed travelling in tar = 5y x /6 x /3 x /2 Ratio of time = : : 3y 4y 5y 1 1 1 : : = 18 12 10 10 15 18 : : = 180 180 180 = 10 : 15 : 18
12 in hour. ∴The clock will show the incorrect time between 1 − 2, 10 − 11, 11 − 12, 12 − 1 day and night both incorrect time (8 × 60) = 480 min Digit l appear in minutes 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51 = 15 min ∴ It will shows the incorrect time = 16 × 15 = 240 min Total incorrect time = 240 + 480 = 720 min Correct time = 24 × 60 − 720 Fraction of correction time 24 × 60 − 720 = 24 × 60 1 = 2
15. (c) Let the student answered correct =x Student answer wrong = y Student unattempted = z According to the question, x + y + z = 30, and 4x + z = 60 x = 15, y = 15, z = 0 x = 14, y = 12, z = 4 x = 13, y = 9, z = 8 x = 12, y = 6, z = 12 x = 11, y = 3, z = 16 x = 10, y = 0, z = 20 Total number of cases = 6
KVPY Question Paper 2011 Stream : SA
132 16. (a) Total length of a pendulum
19. (d) As there is no external force,
23. (c) As total internal reflection occurs
remains same, so extreme point D lies on the line AB, as shown below.
centre of mass of a system remains at same position. Initially,
at angle of incidence, i = 45°. 45° 90° 45°
Box C
D
B
A
This can be proved by applying energy conservation between extreme positions A and D (its given friction is abscent), K A + U A = KB + UB = KD + UD ⇒ ⇒
0 + U A = 0 + UB = 0 + UD U A = UB = UD ⇒ hA = hB = hD
17. (a) Taking boy, toy and ground as a composite system, we can say that there is no external force acting on the system, net acceleration of the system is zero. Asystem = 0 ⇒ (Fnet )system = 0 ⇒ F + (mg )boy + (N )ground + (f )friction = 0
18. (c) Let first ball reaches upto height H and it fells by a distance H − h, where it collided with second ball which rises upto height h. H–h
H h
Equation of motion for first and second ball, 1 …(i) H − h = gt 2 2 1 2 …(ii) h = ut − gt 2 From Eqs. (i) and (ii), we have H u 2 / 2g u H = ut or t = = = u u 2g Substituting the value of t in Eq. (ii), we have 1 u2 u − g× h=u× 2g 2 4g 2 u 2 u 2g − h= 2g 8g 2 h=
u u 4u − u − = 2g 8g 8g 2
or h = 3u 2 / 8 g =
2
2
2
2
3 u 3 × = ⋅H 4 g 4
u QH = g 2
O (20,0)
Position of centre of mass of a system taking girl at origin is 36 × 0 + 9 × 20 9 × 20 = XCM = 36 + 9 45 Finally,
So, using µ = µ=
1 , we have sin C
1 = sin 45°
2 or µ ≈ 142 .
24. (d) When a lens is cut into half or its half part is blackened, image is formed at same place but its intensity is reduced.
25. (c) As material of rod is not changed, Box O
x
20 – x (20,0)
Position of centre of mass when girl and box are at same position is (36 × x) + (9 × x) 36x + 9x = X ′CM = 36 + 9 45 9 × 20 36x + 9x ′ ⇒ As, XCM = XCM = 45 45 ⇒ 9 × 20 = 45x ⇒ x = 4 m So, girl travelled by 4m, when she meet with box.
resistivity of both rods is same. Also, volume of material is same for both rods, so A1 l1 = A2l2 or A1 L = A2 (2L) A A2 = 1 ⇒ 2 l Now, using R = ρ , we have A ρL 2L = 4 R2 = ρ (A1 / 2) A1
26. (b) As direction of fields of charges at points A and B,
and effect of gravity on them is insignificantly low. So, pressure exerted by gas molecules is same everywhere.
22. (c) To reach point Q, using Fermat’s principle, girl must bend her path towards normal as on beach velocity of girl is more than her velocity at sea. Q
EB
EA
EA
20. (a) Most likely each of the object is in thermal equilibrium with its surroundings. So, T1 = T2 = T3 . 21. (b) Gas molecules move randomly
R2 = 4R
or
EB + Q A
EB – 2Q B
are in opposite directions to left of A or right of B, so fields can be zero in these regions. But in right side of B, field cannot be zero as EA is very smaller than EB (charge at A is smaller magnitude and its distance from B is also large). So, field can be zero in region left of A.
27. (b) Useful power Work done by motor Time duration W ×η= ∆t mgh ×η= t 300 × 10 × 6 = 0.4 η= 60 × 750 =
Sea
C
Beach P
So, correct path is PCQ to reach in shortest time. Note Laws of refraction of light follows from Fermat’s principle.
EA
⇒ Pinput ⇒ Pinput ⇒
So, per cent efficiency is, η = 40%
133
KVPY Question Paper 2011 Stream : SA 28. (d) Using Kirchhoff’s junction rule, directions and magnitudes of currents are as, 3A P
Q 3A
6A
1A 4A
10 A
2A
6A 8A 4A
2A
Clearly, current in the portion PQ is from Q to P is 6A.
29. (a) Decay scheme is as given below 214 Pb 82
214 A 84 210 X 82
+ 20 e –1
4 2
+ He
30. (c) From ideal gas equation, pV kBT
Substituting given values, we get 105 × 100 ⇒ N = . × 10−23 × 273 138 ⇒
No reaction It doesn’t give positive iodoform test. (d) CH3 CH2CH2CH2CH2OH + I2 / NaOH
34. (d) Average kinetic energy depends
39. (b) For first order reaction
upon the temperature and not on the type of gases involved. 3kT For any gas, (K. E )avg = per molecule 2 The (K. E)avg of a deuterium molecule is same as that of hydrogen molecule.
35. (a) ∆H for this process = CV ∆T = 0
Following conservation of mass number and atomic numbers, we have Number of neutrons in X is N = A − Z = 210 − 82 = 128 and number of protons is 82. pV = nRT ⇒ pV = NkBT or N =
(i) values of l should range from 0 to n − 1 (ii) values of m should range from −l to l Thus, the set that is not allowed for electron in H atom is n = 2, l = 2, m = − 1 The allowed set of quantum numbers for H  atom having n = 2 will be l = 0 to 1 ml = − 1, 0, 1
N = 3 × 1027 molecules
31. (d) Since the pressure of the helium gas in both the balloons A and B are same. Therefore, the final pressure of He will not change. Hence, the correct option is (d).
32. (c) The electron releasing groups attached to benzene increases the reactivity towards FriedelCrafts alkylation whereas the electron withdrawing groups decreases the reactivity. Among the given groups,—NO2 and COOCH3 are electron withdrawing group EWG, so they will decrease the reactivity where NO2 shows stronger −I effect than COOCH3 . So, the reactivity of nitrobenzene towards FriedelCraft alkylation will be least. OCH3 is an electron donating group, so it will increasing the reactivity. Thus, the order of reactivity towards FriedelCraft alkylation is (iv) > (ii) > (iii) > (i)
33. (d) For any set of principal (n ), azimuthal (l) and magnetic (mi ) quantum numbers, the conditions that are allowed for an electron is
(at constant temperature) ∆S for this process will be positive that is ∆S > 0, the randomness increases the molecules of gases A and B gets intermixed with each other, when the partition is removed. Thus, the correct option is (a).
36. (d) The thermal decomposition of (NH4 )2 Cr2O7 gives chromium oxide (Cr2O3 ), nitrogen gas and water. ∆
(NH4 )2 Cr2O7 → Cr2O3 + N2 + 4H2O
37. (a) From the graph it can be seen that solubility of KNO3 in water at 40°C is approximately 200 g per 100 of water. ∴Amount of KNO3 that dissolve (or solubility) in 50 g of water will 200 = × 50 = 100 g 100
38. (a) Iodoform test with sodium hypoiodite is used for the detection of CH3 CO group or CH3 CH(OH) group which produces CH3 CO group on oxidation. Iodoform reaction with the given compounds are as follows : (a) CH3 —CH2— CH2— C— CH3 + I2/  NaOH O 2pentanone
H + / H 2O
→ CH3 CH2CH2COOH + CHI3 Butanoic acid
Iodoform
OH Η +/Η 2Ο CH3 CH2 C HCH2CH3 + I2 / ΝaΟΗ →
(c)
3pentanol
1pentanol H + / H 2O
→ no reaction It also give negative iodoform test. k=
a 2. 303 log t a − x
also k =
0.693 t1/ 2
a 0.693 2. 303 = log t1/ 2 t a − x
∴
According to question, 0.693 2. 303 a = log 2 × 60 t1/ 2 a / 16 ⇒
t1/ 2 = 30 min
40. (c) In the process of roasting, sulphide ore is converted into an oxide ore with a regular supply of air in a furnace at a temperature below the melting point of the metal. Thus, the conversion of sulphide ore into metal oxide is given the reaction ZnS → ZnO, hence option (c) is correct.
41. (d) The structures of given compounds can be drawn as follows : O
O
O
P
P
P
OH , H H3PO2
H
H, OH H3PO3
HO
OH OH H3PO4
HO
Thus, the number of P– H bond(s) in H3 PO2 , H3 PO3 and H3 PO4 respectively are 2, 1, 0.
42. (d) When chlorine gas is passed through an aqueous solution of KBr, the solution turns orange brown due to the evolution of bromine gas. The equation for the above reaction can be written as Cl 2 + 2KBr → 2KCl + Br2 ↑
Orange brown
43. (b) The conditions for a compound to
If gives positive iodoform test due to the presence of CH3 CO group. (b) CH3 —CH2— C—CH2— CH3 + I2/NaOH  O
be aromatic are (i) the molecule should be planar. (ii) it should be cyclic with alternate single and double bonds.
→ no reaction It gives negative iodoform test.
(iii) it should follows Huckel’s rule, i.e. should have (4n + 2) π electrons.
3pentanone H + / H O 2
KVPY Question Paper 2011 Stream : SA
134 π electrons present in given compounds are as follows : H N (i)
(4n+2)π electrons = 6π electrons
∴ Follows Huckel’s rule (4n+2)π electrons = 8π electrons
(ii)
This reaction involves SN 2 mechanism where CN − is substituted over Cl − as it is a good leaving group as compared to Br and I, and occurs at a primary carbon (sp3 hybridised). 46. (b) Attached earlobes is an autosomal recessive trait. Thus, a heterozygous parent with free earlobes will have offspring with attached earlobes. Free earlobes
Free earlobes
×
Aa
+
Aa
Aa
aa
Aa
Parents
Doesn’t follow Huckel’s rule. AA
(4n+2)π electrons = 10π electrons
(iii)
Free earlobes
47. (d) For a complete meiotic cell
Follows Huckel’s rule. O
division, there takes place one round of DNA replication during the Sphase and two divisions. This is because meiosis is a process where a single cell divides twice to produce four cells containing half the original amount of genetic information.
(4n+2)π electrons = 6π electrons
(iv) N
Follows Huckel’s rule. Thus, compound (ii) is not aromatic.
44. (b) The IUPAC nomenclature of the structures given in the options are as follows 1
2
H3C
3
(i)
5
Injury in blood vessels Platelets clump at the wound
CH3
H 3C
Platelets release thrombokinase
5 6
(ii)
48. (c) Blood clotting involves the conversion of fibrinogen to fibrin. The blood clotting mechanism takes place as follows
3, 4dimethyl hexane
CH3
4
6
Offsprings Attached earlobes
4
H 3C
1
3
CH3
2
2, 3dimethyl hexane
Thrombin
Thrombokinase
VitaminK
H 3C
Fibrin
CH3
1
H3C 2
(iii)
3, 4dimethyl hexane
5
3 4
6
CH3
H3C
Prothrombin
Fibrinogen
Clot forms to prevent further blood loss
49. (b) The gall bladder is involved in
Hence, the correct option is (b).
storing and secreting bile. The gall bladder is a pearshaped, hollow structure located under the liver and on the right side of the abdomen. Its primary function is to store and concentrate bile, a yellow brown digestive enzyme produced by the liver.
45. (c)
50. (c) Green light is not at all useful for
CH3 (iv) H 3C 1
CH3
CH3
2
4 3
2, 4dimethyl hexane
CH3
6 5
photosynthesis. This is because, plant reflects green light and due to this same reason, plants appear green in colour.
Br Cl
CN – DMSo, ∆
51. (d) Tidal volume is the volume of
Br
inspired/expired air moving in and out of the lungs with each breath.
I CN
I
Vital capacity is the volume that can be inspired/expired after full expiration/inspiration.
Total lung capacity refers to the total amount of air in the lungs after taking the deepest breath possible.
52. (a) A normal human inherits only one sex chromosome (either X or Y chromosome) from father. Sex chromosomes, determine whether an individual is male or female. In human and other mammals these are designated by scientists as X and Y. In humans, the sex chromosomes comprise one pair of the total of 23 pairs of chromosomes. The other 22 pairs of chromosomes are called autosomes. 53. (c) Scurvy is a result of the deficiency of vitaminC. VitaminC is mainly found in fruits such as oranges, grapefruit, lemons, strawberries and melons or it is found in vegetables such as broccoli and bell peppers. Therefore, malnutrition causes vitaminC deficiency. Protein deficiency malnutrition is known as kwashiorkor and marasmus. VitaminD deficiency causes rickets. Carbohydrates deficiency causes weakness, nausea, dehydration, etc.
54. (c) Centriole is not found in plant cells, it is found only in animal cells. These paired organelles are typically located together near the nucleus in the centrosome, a granular mass that serves as an organising centre for microtubules. Centriole is involved in the development of spindle fibres in cell division.
55. (a) The brain and spinal cord are made up of many cells, including neurons and glial cells. Neurons are cells that send and receive electrochemical signals for pain or pleasure to and from the brain and nervous system. l Blastocyst is a structure formed in the early development of mammals. It possesses an Inner Cell Mass (ICM) which subsequently forms the embryo. l Histoblast is a cell or cell group possessing broad histogenetic capacity, i.e. capable of forming tissue. l Haemocyte is a cell of the haemolymph of various invertebrates, especially arthropods. 56. (b) Benedict’s test is used to detect reducing sugars such as glucose. Sucrose is a nonreducing sugar, it gives negative result for Benedict’s test. Benedict’s reagent is a complex mixture of sodium carbonate, sodium citrate and copper (II) sulphate petahydrate.
135
KVPY Question Paper 2011 Stream : SA 57. (a) Iodine is found in thyroxine. Thyroxine, also called 3, 5, 3′ , 5′tetraiodothyronine or T4, is one of the two major hormones secreted by the thyroid gland (the other is triiodothyronine). Thyroxine’s principal function is to stimulate the consumption of oxygen and thus the metabolism of all cells and tissues in the body. Thyroxine is formed by the molecular addition of iodine to the amino acid tyrosine while the latter is bound to the protein thyroglobulin.
∴
f (x) = 2x2 − 5x + 3 f (50) = 2(50)2 − 5(50) + 3 = 5000 − 250 + 3 = 4753 Now, 1000t < f (50) < 1000(t + 1) ∴ 1000t < 4753 < 1000(t + 1) ⇒ t < 4.753 < t + 1 ∴ t = 4 , t is integer.
62. (c) Let S=
(2011) + 1 2 +1 3 +1 4 +1 + ... + + + (2011)2 − 1 22 − 1 32 − 1 42 − 1 2
2
2
r2 + 1
Here,
Tr =
⇒
Tr =
r2 −1
58. (b) Lactometer is a device used for testing the purity of milk. It measures relative density of milk with respect to water, which is also called specific gravity. If the specific gravity of a sample of milk is within the approved ranges, the milk is pure. If it is not, then there is some adulteration in milk.
59. (b) When mammalian liver cells are kept in a hypotonic solution, endosmosis occurs as the cell is hypertonic. Due to the endosmosis, the cellular protoplasm is filled with water, it swells and the cells become turgid. Swelling is seen because the water flows from lower concentration of solute to the higher concentration of solute.
⇒ ⇒
r2 −1 + 2
r2 −1 2 2 = 1+ 2 = 1+ (r − 1)(r + 1) r −1 1 1 Tr = 1 + − r −1 r + 1 S= =
2011
∑ Tr
r= 2 2011
1
1
∑ 1 + r − 1 − r + 1
r= 2
⇒
2
S = T2 + T3 + T4 + ... + T2011
1 1 1 1 ⇒ S = 1 + − + 1 + − 1 3 2 4
of cancer that develops from epithelial cells. Lymphoma is the cancer that occurs in lymph cells. Leukemia is blood cancer that originates in the blood and bone marrow. Mesoderm is one of the germ layer from which skeletal muscle, bone, connective tissue, heart and the urogenital system originate.
1 1 1 1 + 1 + − + ... + 1 + − 3 5 2010 2012 1 1 1 − ⇒ S = 2010 + 1 + − 2 2012 2011 1 1 1 + ⇒ S = 2011 + − 2 2011 2012 1 ⇒ S is lie between 2011, 2011 . 2
61. (c) We have,
63. (b) Given,
f (x) = ax2 + bx + c, a , b, c,∈ Z Also f (1) = 0, 40 < f (6) < 50, 60 < f (7) < 70 ∴a + b + c = 0, 40 < 36a + 6b + c < 50, 60 < 49a + 7b + c < 70 …(i) c= −a− b ∴ 40 < 36a + 6b − a − b < 50 and 60 < 49a + 7b − a − b < 70 ⇒ 40 < 35a + 5b < 50 and 60 < 48a + 6b < 70 70 ⇒ 8 < 7a + b < 10 and 10 < 8a + b < 6 Now, a and b are integer ∴ 7a + b = 9 and 8a + b = 11 On solving these equation, we get a = 2, b = − 5 Put the value of a , b in Eq (i), we get c=3
Diameter of base = 100 mm 100 mm = 50mm = 5 cm ∴Radius of base = 2 Let other radius of base = r And height of truncated cone = h Volume of initially truncated cone = V πh ∴ V = {(5)2 + 5r + r 2 } 3 When radius increase by 21% 605 21 ×5= ∴Radius of base = 5 + 100 100 When volume increase by 21% 21 V 121 V Then, V 1 = V + = 100 100 2 πh 605 605 Now, V 1 = r + r 2 + 3 100 100
60. (d) Carcinoma is a category of types
[Q r and h are same]
121 V πh (605)2 + 60500r + (100r )2 = 100 3 (100)2 121 πh 2 ⇒ × (25 + 5r + r ) 100 3 πh (605)2 + 60500r + (100r )2 = 3 10000
⇒
⇒100(3025 + 605r + 121r 2 ) = 366025 + 60500r + (100r )2 ⇒100 × 121r 2 − 10000r 2 = 366025 − 302500 ⇒ 2100r 2 = 63525 63525 = 30.25 r2 = ⇒ 2100 r = 30.25 = 5.5 cm = 55 mm ⇒
64. (d) Let the speed of B = x km/h and the speed of A = 3x km/h Distance between A and B = 30 km Given, distance between them decrease at 2 km per minutes. ∴ Distance decrease in one hour = 2 × 60 = 120 km ∴Total distance travelled by A and B in one hour = (x + 3x) km = 4x km 120 = 30 km/h ∴Speed = 4 Hence, speed of B = 30 km/h Speed of A = 90 km/h Distance travelled by A and B after 10 min = 2 × 10 = 20 km So, remaining distance = (30 − 20) = 10 km Time taken by B to distance travelled 10 10 km = × 60 = 20 min 30 Total time taken by A = 20 + 10 = 30 min
65. (a) Taps A , B ,C fill up a tank independently 10 h, 20 h, 30 h, respectively. Given, exactly one pairs of taps is open during each hour and every pairs of taps is open at least one hour. First A and B are open for one hour, then B and C and then C and A 1 1 1 1 1 1 + + + ∴ + + 10 20 20 30 30 10 1 1 1 12 + 6 + 4 + + = 60 5 10 15 22 11 = = 60 30 In three hours the tank will be filled 11 th part. Now, for minimum time, 30 the rest of tank must be filled with A and B taps. =
KVPY Question Paper 2011 Stream : SA
136 1 + 1 = 30 10 20 20 11 19 So, the rest of 1 − th = th part 30 30 of tank will taks 5 h more. So, the tank will be filled in 8 h.
∴
66. (b) For equilibrium of block hung
Now, by gas equation, we have As, n1 + n2 = n 4V × 0.5 5V × p1 V ⇒ + = RT RT RT ⇒ p = 0.6 atm
69. (c) Ray diagram for pool is as shown below.
from string, Spring force + Buoyant force = Weight of block
r h
m
x/2
ρVg
So, we have and
kx1 + ρ1Vg = ρVg
…(i)
kx2 + ρ2Vg = ρVg
…(ii)
Eliminating k, we get ρ x − ρ2x1 ρ= 1 2 x2 − x1
67. (d) By workkinetic energy theorem, work done is equal to change in kinetic energy. x=8
∫x = 4 Fdx = ∆K
So,
F 3 1.5
4
8 10
x
⇒ Area under forcedisplacement graph 1 m (vf2 − vi2 ) 2 1 1 1 (3 − 15 . ) × (8 − 4) = × ⇒ 2 2 2
from x = 4 to x = 8 =
× (vf2 − (3.16)2 ) −1
⇒
vf = 6. 8 ms
p=1 atm V T
p=0.5 atm 4V T
Let final temperature after opening the valve is Tf , then ∆Wext = 0 and ∆Qext = 0 So, from first law of thermodynamics, ∆U = 0 ⇒ n1CV T + n2CV T = (n1 + n2 ) CV Tf ⇒
Tf = T
Using n1 ⋅ sin i = n2 ⋅ sin r , we have 4 …(i) 1 × sin (90° − θ) = sin r 3 x 4 2 Also, = = tan r = 2h 7 × 2 7 2 sin r = ⇒ 53 Substituting sin r in Eq. (i), we have 4 2 8 cos θ = × = 3 53 3 53 70. (a) From given circuit, if i = circuit current, then 10 V = iReq ⇒ i = = 2.5 A 4 So, power dissipated by circuit is 25 W P = i 2R = (2.5)2 × 1 = 4 When 1Ω resistor is replaced by a 9Ω resistor, then power dissipated in 9 Ω resistor is 2 10 P ′ = × (9) 12 100 × 9 25 W = = 12 × 12 4 So, P ′ = P
,
,
,
an alkene is getting converted into epoxy group, so an oxidising agent is required both H2O2 and C6 H5 COOH are oxidising agent but C6 H5 COOH is used as they are not very sensitive to solvent polarity while, in 2nd sequence of reaction, the epoxy group is being reduced into an alcoholic group, thus a reducing agent is required. Thus, the suitable reagents are
According to Henderson equation [salt] pH = pK a + log [acid] log [CH3 COO− ] pH = pK a + log [CH3 COOH] 10 pH = 4.76 + log 10 pH = 4.76 + log 1 pH = 4.76 ⇒
72. (c) Nine structure isomers are possible for the hydrocarbon having molecular formula C 4 H 6 .
O
C6H5COOH oxidation (Peroxide attack)
(Epoxy group) LiAlH4 Reduction
OH
74. (a) The magnetic character of the given complexes are as follows : (i) Co(NH 3 )6 ]3 + Oxidation state of Co in [Co(NH3 )6 ]3+ is +3. The electronic configuration of Co3 + is 3d 6 4s0 . 3d
4s
4p
As NH3 is a strong field ligand, paring of electrons occur [Co(NH3)6]3+
3d ××××
71. (d) Meq of CH3COOH = 100 × 0.1 = 10 Meq of CH 3 COONa = 50 × 0.2 = 10
68. (b) Given situation is
,
73. (b) In the first sequence of reaction,
i θ
ρf Vg
kx
These are as follows : CH 2 == C == CH — CH 3 , CH 3 CH 2 C ≡≡ CH, CH 3 — C == C — CH 3 , CH 2 == CH — CH == CH 2
4s ××
4p ×× ×× ××
d 2sp 3 hybridised
Thus, the complex is diamagnetic. (ii) [Ni(NH 3 )6 ]2 + Oxidation state of Ni in [Ni (NH3 )6 ]2+ is +2. The electronic configuration for Ni 2 + is 3d 8 4s0 . 3d
4s
4p
5d
137
KVPY Question Paper 2011 Stream : SA Though NH 3 is a strong ligand pairing will not occur because, if pairing would occur then also 2,dorbitals will not be available for hybridisation. 3d [Ni(NH3)6 4s ××
4p
5d ××××
sp3d2
Thus, the complex is paramagnetic. (iii) [Cr(H 2O)6 ]3 + Oxidation state of Cr in [Cr (H2O6 )]3+ is + 3. The electronic configuration of Cr 3 + is [Ar] 3d3 4s0 . [Cr(H2O)6]3+ 3d
4s ××
×× ××
4p ×××× ××
d2sp3
As H 2O is a weak field ligand pairing of electrons will not occur and the complex is paramagnetic. (iv) [Fe(H 2O)6 ]2 + Oxidation state of Fe in [Fe (H2O6 )]2+ is + 2. The electronic configuration of Fe 2 + is [Ar] 3d 6 4s0 . 3d [Fe(H2O6)]2+ 4p ×× ×× ×× sp3d2
5d ×× ××
4s ××
75. (a) For the reaction, H 2 ( g ) + I2 ( g ) r 2HI( g ) [KC = 46] [HI]2 QC = [H 2 ] [I2 ] 0.4 × 0.4 ⇒ QC = 8 QC = 01 . × 0.2
]2+
××××××
As H 2O is a weak field ligand, pairing of electrons will not occur and thus the complex will be paramagnetic.
As QC < KC So, the reaction will proceeds in forward direction. Hence, amount of HI increases.
76. (b) Tobacco is a dicot plant, thus its leaves have more number of stomata on its lower surface. If you cover the leaves of a healthy plant with vaseline, it will block its stomata and therefore it will not lose water through transpiration, so the upward movement of the water in the plant will stop. This will not allow the plant to wilt quickly. The leaf ‘2’ is smeared with vaseline on the upper surface, so the plant will lose water from the lower surface and in leaf ‘3’, vaseline is smeared on the lower surface therefore the water is lost from the upper surface. But the number of stomata are more on the lower surface therefore leaf ‘2’ will wilt more quickly than leaf ‘3’.
77. (d) Vestigial organs are those organs which are no longer in use. They were used to play an important role in our ancestors, but as and when we developed and evolved, some of these organs lost their functionality but managed to stay in our body. Appendix, coccyx, external
ears, etc. are some examples of vestigial organs in human body.
78. (c) Gregor Mendel studied inheritance of traits in pea plants. In a monohybrid cross, the recessive trait disappears in the first filial generation. The traits that were visible in the F1generation are referred to as dominant traits. This happens because recessive allele does not express itself in the presence of dominant allele. 79. (a) An individual gets one sex chromosome from each parent during fertilisation. +
XX
XY
XX XY
XX XY
Girl
Girl
Boy
Boy
We see from the above cross that sons get their Ychromosome from father and Xchromosome from mother. Therefore, if a man with an Xlinked dominant disease has six sons with a woman having a normal complement of genes, their sons will show no symptom of the disease.
80. (c) In evolutionary terms, an Indian school boy is more closely related to Chinese horse. Human and horse irrespective of their country are more generally similar in their chromosomal management, than to rest of the species mentioned here. This is because human and horse both belong to class Mammalia. Other options like frog, snake and shark belong to class– Amphibia, Reptilia and Chondrichthyes.
KVPY Question Paper 2010 Stream : SA
138
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2010 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) 5. The sides of a ∆ ABC are positive integers. The
MATHEMATICS 1. A student notices that the roots of the equation x2 + bx + a = 0 are each 1 less than the roots of the equation x2 + ax + b = 0. Then, a + b is (a) possibly any real number (b) −2 (c) −4 (d) −5
2. If x, y are real numbers such that 3( x / y ) + 1 − 3( x / y ) − 1 = 24, (b) 1
(c) 2
(d) 3
3. The number of positive integers n in the set {1, 2, 3, ………, 100} for which the number 12 + 22 + 32 + ..... + n 2 is an integer is 1 + 2 + 3 +.... + n (a) 33
(b) 34
(c) 50
(d) 100
4. The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths 39, 40, 41. The length of the main diagonal of the cuboid which joins a pair of opposite corners is (a) 49
(b) 49 2
(a) The area of ∆ABC is always a rational number (b) The area of ∆ABC is always an irrational number (c) The perimeter of ∆ABC is an even integer (d) The information provided is not sufficient to conclude any of the statements A , B or C above
6. Consider a square ABCD of side 12 and let M , N be
then the value of (x + y)/(x − y) is (a) 0
smallest side has length 1. Which of the following statements is true?
(c) 60
(d) 60 2
the midpoints of AB, CD respectively. Take a point P on MN and let AP = r , PC = s. Then, the area of the triangle whose sides are r , s,12 is (a) 72
(b) 36
(c)
rs 2
(d)
rs 4
7. A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a m by a rope of length 5a m in a grass field. (The cow cannot graze inside 2 the fenced area). What is the maximum possible area of the grass field to which the cow has access to graze? (a) 5 πa 2
(b)
5 2 πa 2
(c) 6 πa 2
(d) 3 πa 2
139
KVPY Question Paper 2010 Stream : SA 8. A closed conical vessel is filled with water fully and is placed with its vertex down. The water is let out at a constant speed. After 21 min, it was found that the height of the water column is half of the original height. How much more time in minutes does it require to empty the vessel? (a) 21
(b) 14
(c) 7
(d) 3
9. I carried 1000 kg of watermelon in summer by train. In the beginning the water content was 99%. By the time I reached the destination, the water content had dropped to 98%. The reduction in the weight of the watermelon was (a) 10 kg
(b) 50 kg
(c) 100 kg
(d) 500 kg
10. A rectangle is divided into 16 subrectangles as in the figure, the number in each subrectangle represents the area of that subrectangle. What is the area of the rectangle KLMN ? 10
M 12
K
(a) 50 kg
(b) 75 kg
(c) 100 kg
(d) 125 kg
14. A planet of mass is moving around a star of mass M and radius R in a circular orbit of radius r. The star abruptly shrinks to half its radius without any loss of mass. What change will be there in the orbit of the planet? (a) The planet will escape from the star (b) The radius of the orbit will increase (c) The radius of the orbit will decrease (d) The radius of the orbit will not change
which P , Q, R and S are fixed resistances, G is a galvanometer and B is a battery. For this particular case, the galvanometer shows zero deflection. Now, only the positions of B and G are interchanged, as shown in figure (ii). The new deflection of the galvanometer
15 25
(a) 20
alrplane gets detached from his parachute. He is able to resist the downward acceleration, if he shoots 40 bullets a second at the speed of 500 m/s. If the weight of a bullet is 49 g, what is the weight of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity, g = 9.8 ms−2.
15. Figure (i) below shows a Wheatstone’s bridge in
4
N
13. A soldier with a machine gun, falling from an
L
(b) 30
(c) 40
(d) 50
Q
P
Q
P
PHYSICS
B
G
11. A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at a constant rate. Which of the following statements describes the variation of the time period T of the pendulum as the water flows out? (a) T decreases first and then increases (b) T increases first and then decreases (c) T increases throughout (d) T does not change
A steadily increasing horizontal force is applied such that the block starts to slide on the table without toppling. The force is continued even after sliding has started. Assume the coefficients of static and kinetic friction between the table and the block to be equal. The correct representation of the variation of the frictional force f, exerted by the table on the block with time t is given by f f (b)
(0, 0)
t
(0, 0)
t
(0, 0)
t
(0, 0)
Figure (ii)
(a) is to the left (b) is to the right (c) is zero (d) depends on the values of P ,Q , R and S
16. 12 positive charges of magnitude q are placed on a circle of radius R in a manner that they are equally spaced. A charge Q is placed at the centre, if one of the charges q is removed, then the force on Q is (a) zero qQ (b) away from the position of the removed charge 4 πε0 R 2 11qQ away from the position of the removed charge (c) 4 πε0 R 2 qQ (d) towards the position of the removed charge 4 πε0 R 2
under 220 V, consuming 1 kW power. Part of its coil burned out and it was reconnected after cutting off the burnt portion.The power it will cunsume now is
(d)
(c)
G
17. An electric heater consists of a nichrome coil and runs
f
f
R
S
B
Figure (i)
12. A block of mass M rests on a rough horizontal table.
(a)
R
S
t
(a) more than 1 kW (c) 1 kW
(b) less than 1 kW, but not zero (d) 0 kW
KVPY Question Paper 2010 Stream : SA
140 18. White light is split into a spectrum by a prism and it is seen on a screen. If we put another identical inverted prism behind it in contact, what will be seen on the screen ? (a) Violet will appear where red was (b) The spectrum will remains same (c) There will be no spectrum, but only the original light with no deviation (d) There will be no spectrum, but the original light will be laterally displaced
19. Two identical blocks of metal are at 20°C and 80°C, respectively. The specific heat of the material of the two blocks increases with temperature. Which of the following is true about the final temperature T f when the two blocks are brought into contact (assuming that no heat is lost to the surroundings)? (a) Tf will be 50°C (b) Tf will be more than 50°C (c) Tf will be less than 50°C (d) Tf can be either more than or less than 50° C depending on the precise variation of the specific heat with temperature
20. A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature tx in this scale is related to the absolute temperature T by tx = 3T + 100. If the specific heat of a material using this unit is 1400 J kg −1K −1, its specific heat in the SI system of units is (a) (b) (c) (d)
4200 J kg −1 K−1 1400 J kg −1 K−1 466.7 J kg −1 K−1 impossible to determine from the information provided
25. Among NH3 , BCl3 , Cl2 and N2, the compound that does not satisfy the octet rule is
(a) NH3
(b) BCl3
(c) Cl 2
(d) N2
26. The gas produced on heating MnO2 with conc. HCl is (a) Cl 2
(b) H2
(c) O2
(d) O3
27. The number of covalent bonds in C 4 H7 Br , is (a) 12
(b) 10
(c) 13
(d) 11
28. An aqueous solution of HCl has a pH of 2.0. When water is added to increase the pH to 5.0, the hydrogen ion concentration (a) remains the same (c) increases threefold
(b) decreases threefold (d) decreases thousandfold
29. Consider two sealed jars of equal volume. One contains 2 g of hydrogen at 200 K and the other contains 28 g of nitrogen at 400 K. The gases in the two jars will have (a) the same pressure (b) the same average kinetic energy (c) the same number of molecules (d) the same average molecular speed
30. Identify the stereoisomeric pair from the following choices. (a) CH 3 CH 2 CH 2OH and CH 3 CH 2OCH 3 (b) CH 3 CH 2 CH 2 Cl and CH 3 CHClCH 3
H  (c) CH3 C == C CH3 and CH3 C C CH3    H H H CH2 and
(d)
CHEMISTRY 21. The boiling point of 0.01 M aqueous solutions of sucrose, NaCl and CaCl 2 would be (a) the same (b) highest for sucrose solution (c) highest for NaCl solution (d) highest for CaCl 2 solution
22. The correct electronic configuration for the ground state of silicon (atomic number = 14) is
(a) 1s2 2s2 2 p 6 3s2 3 p 2 (c) 1s2 2s2 2 p 4 3s2 3 p 4
(b) 1s2 2s2 2 p 6 3 p 4 (d) 1s2 2s2 2 p 6 3s1 3 p5
23. The molar mass of CaCO3 is 100 g. The maximum amount of carbon dioxide that can be liberated on heating 25 g of CaCO3 is
(a) 11 g (c) 22 g
(b) 5.5 g (d) 2.2 g
24. The atomic radii of the elements across the second period of the periodic table (a) decrease due to increase in atomic number (b) decrease due to increase in effective nuclear charge (c) decrease due to increase in atomic weights (d) increase due to increase in the effective nuclear charge
BIOLOGY 31. Which of the following is a water borne disease? (a) Tuberculosis (c) Chicken pox
(b) Malaria (d) Cholera
32. In his seminal work on genetics, Gregor Mendel described the physical traits in the pea plant as being controlled by two ‘factors’. What term is used to define these factors today? (a) Chromosomes (c) Alleles
(b) Genes (d) Hybrids
33. A majority of the tree species of peninsular Indian origin fruit in the months of (a) AprilMay (c) DecemberJanuary
(b) AugustSeptember (d) All months of the year
34. In frogs, body proportions do not change with their growth. A frog that is twice as long as another will be heavier by approximately (a) twofold (b) fourfold (c) sixfold
(d) eightfold
141
KVPY Question Paper 2010 Stream : SA 35. Which of the following has the widest angle of binocular vision? (a) Rat
(b) Duck
(c) Eagle
(d) Owl
36. The two alleles of a locus which an offspring receives from the male and female gametes are situated on (a) two different homologs of the same chromosome (b) two different chromosomes (c) sex chromosomes (d) a single chromosome
37. Ants locate sucrose by (a) using a strong sense of smell (b) using a keen sense of vision (c) physical contact with sucrose (d) sensing the particular wavelength of light emitted/reflected by sucrose
38. The interior of a cow dung pile kept for a few days is quite warm. This is mostly because (a) cellulose present in the dung is a good insulator
(b) bacterial metabolism inside the dung releases heat (c) undigested material releases heat due to oxidation by air (d) dung is dark and absorbs a lot of heat
39. Which one of these is the correct path for a reflex action? (a) Receptor → Motor neuron→ Spinal cord → Sensory neuron→ Effector (b) Effector→ Sensory neuron→ Spinal cord → Motor neuron→ Receptor (c) Receptor→ Sensory neuron→ Spinal cord→ Motor neuron→ Effector (d) Sensory neuron → Receptor→ Motor neuron→ Spinal cord→ Effector
40. Insectivorous plants digest insects to get an essential nutrient. Other plants generally get this nutrient from the soil. What is this nutrient? (a) Oxygen (c) Carbon dioxide
(b) Nitrogen (d) Phosphates
PARTII (5 Marks Questions) MATHEMATICS 1. In a ∆ ABC , D and E are points on AB, AC respectively such that DE is parallel to BC. Suppose BE , CD intersect of O. If the areas of the triangles ADE and ODE are 3 and 1 respectively. Find the area of the ∆ ABC, with justification.
2. Leela and Madan pooled their music CD’s and sold them. They got as many rupees for each CD as the total number of CD’s they sold. They share the money as follows: Leela first takes 10 rupees, then Madan takes 10 rupees and they continue taking 10 rupees alternately till Madan is left out with less than 10 rupees to take. Find the amount that is left out for Madan at the end, with justification.
3. (a) Show that for every natural number n relatively prime to 10, there is another natural number m all of whose digits are 1’s such that n divides m. (b) Hence or otherwise show that every positive rational number can be expressed in the form a for some natural numbers a , b, c. b 10 (10c − 1)
A RA
R V RV
+ – P
R V RV + –
A RA
Q
In each case, the resistance is estimated by using V Ohm’s law Rest = , where V and I are the readings I of the voltmeter and the ammeter, respectively. The meter resistances RV and RA are such that RA s2. s1 >1 ⇒ s2 (Tf − Ti )Cold block ⇒ >1 (Ti − Tf )Hot block ⇒
T − 20 >1 80 − T
⇒
T − 20 > 80 − T
⇒
T + T > 80 + 20
⇒
2T > 100
⇒
T > 50°C
So, change in temperatures in two systems are related as tx 2 − tx1 = (3T2 + 300) − (3T1 + 300) = 3 (T2 − T1 ) or ∆tx = 3∆T Now to convert units, we use N1 u1 = N 2u2 N2 J 1400 J = ∆tx kg ∆T kg 1400 N 2 = ⇒ 3 ∆T ∆T 1400 ∆T ⇒ N2 = 3∆T 1400 = 466.7 N2 = ⇒ 3 Hence, in SI system, value of specific heat is c = 466.7 J kg −1 K −1 .
21. (d) As boiling point is a colligative property, it depends only on the number of particles present in a solution. Aqueous solution having more number of particles will have highest elevation in boiling point. C12H22O11 (aq) r No ions NaCl r Na + + Cl − (Total ions = 2) CaCl 2 r Ca 2+ + 2Cl − (Total ions = 3) Thus, CaCl 2 will have highest boiling point.
22. (a) As silicon belongs to group 14 whose general configuration is ns2np 2. Thus, the correct electronic configuration of Si is 1s2 2s2 2 p 6 3s2 3 p 2.
respective octets (i.e., 8 electrons in their outermost shell or 2 electrons in case of H, Li and Be to attain stable nearest noble gas configuration). The Lewis structures of given molecules are as follows (a) NH3
H
H
As the octet is complete, hence it follows octet rule.
(b)
×× × × Cl× × × ×× ×× BCl3, ×× Cl× B × Cl ×× ×× ××
Octet rule is not followed in this case as there are total 6 electrons in the outermost shell of B even after bonding. ×× × Cl× × ××
(c) Cl2 , Cl
Octet rule is followed in case of Cl 2 . (d) N2 ,
× × ×
N
N
As both the N have complete octet. Thus, it follows octet rule. Hence, the correct option is (b).
26. (a) On heating MnO2 with conc. HCl, MnCl 2 and H 2O are released with the evolution of Cl 2 gas. The chemical equation for it can be written as MnO2 + 4HCl → MnCl 2 + 2H2O + Cl 2
27. (a) The structure of C4 H7 Br can be written as H
∆
23. (a) CaCO3 → CaO + CO2 25 1 = mole 100 4 1 mole of CaCO3 gives 1 mole of CO2. 1 1 ∴ mole of CaCO3 gives mole of CO2. 4 4 1 mole of CO2 = 44 g 44 1 So, mole of CO2 = = 11 g 4 4 Thus, 11 g of CO2 is released on heating 25 g of CaCO3 .
×
H× N
H
1
2
C
H 3 4
C
H 5 6
No. of moles of CaCO3 =
24. (b) On moving across the second
H
7
C
9
8
H
2 10 12
C
Br
11
H
Thus, total number of covalent bonds in C 4 H 7Br is 12.
28. (d) As we know pH = − log [H + ] As the initial pH of aqueous solution = 2 ∴ Concentration of H+ , [H + ] = 10−2 As the final pH of aqueous solution = 5 [H+ ] = 10−5
∴
[H+ ]f
10−5
period of the periodic table, the atomic radii of the elements decrease, due to increase in effective nuclear charge, the electrons of all the shells are pulled little closer to nucleus thereby making each individual shell smaller and smaller.
∴
25. (b) According to octet, the atoms of
29. (c) For Ist jar
different elements combine with each other in order to complete their
No. of moles of H 2 =
+
[H ]i
=
10−2
= 10−3 +
Thus, H concentration decreases by thousand fold. 2 =1 2
147
KVPY Question Paper 2010 Stream : SA ∴No. of molecules of H 2 in 1 mole = 6.022 × 1023 For 2nd jar 28 No. of moles of N 2 = =1 28 ∴No. of molecules of N2 in 1 mole = 6.022 × 1023 Thus, the gases in two jars will have same number of molecules.
30. (c) The type of isomers in the given options are as follows (a) CH3 CH2CH2OH and CH3 CH2OCH3 They are functional isomers. (b) CH3 CH2CH2Cl and CHCHClCH3 They are positional isomers. (c) H3C
C
CH3
C
H
H Cis
H and
CH3C
C
CH3
H Trans
As they are cis and trans form, so they are geometrical isomers which are a type of stereoisomers. CH2 (d)
and
These are structural isomers. Thus, the correct option is (c).
34. (d) A frog that is twice as long as another will be heavier by approximately eight folds. It is because their weight is determined by their folds, so by applying unitary method it is eight folds. Frog shape is triangular.
13
13
13 =1 Initial size
23
23
23 =8 Final size
35. (d) Owl’s eye have the widest angle of binocular vision among rat, duck and eagle. Binocular vision is the vision using two eyes with overlapping fields of view, allowing good perception of depth. Unlike many birds with eyes that sit at an angle, owl’s eye face directly forward giving them incredible binocular vision. Although their large eyes cannot move or rolllike human eyes can, owls can move their heads nearly all the way around, allowing them to have a 270 degree range of vision without moving their bodies. 36. (a) An allele is a varient form of gene. Some genes have a variety of different forms, which are located at the same position on a chromosome. Humans are called diploid organisms because they have two alleles at each genetic locus, with one allele inherited from each parent. Each pair of alleles represents the genotype of a specific gene.
32. (c) The term allele was formerly used by Mendel for the factors representing the two alternate forms of a character, e.g. tallness and dwarfness in case of height in pea (T and t).
37. (c) Ants locate sucrose by physical contact with sucrose. Ants can smell food using their antennae, which can detect minute odours. But sugars actually do not have a smell. Therefore, when the scout worker ants locate sugar while foraging, they take a piece back home to the colony, while doing this, the ant will leave a chemical trial which the other ants can follow.
33. (a) This happens because the
38. (b) The interior of a cow dung pile
temperature starts to increase all over the country in March and by April. The interior parts of the peninsular record mean daily temperature of 3035°C. Maximum temperature rises sharply exceeding 45°C by the end of May. The warm temperature can affect the ripening process of fruits and vegetables by speeding up the production of ethylene gas which rushes the ripening or fruiting process.
kept for a few days is quite warm. This is because bacterial metabolism inside the dung releases heat in the form of biogas (CH4 + CO2 ). This happens by anaerobic digestion of cow dung by methanogenic bacteria.
31. (d) Water borne diseases such as cholera is caused by drinking contaminated or dirty water. Cholera is caused by the bacterium Vibrio cholerae.
39. (c) The correct path for a reflex action is Receptor → Sensory neuron → Spinal cord → Motor neuron → Effector
The sensory nerve fibres bring sensory impulses from the receptor organ to the central nervous system (brain and spinal cord). The motor nerve fibres relay the motor impulses from the central nervous system to the effector organs. Reflex action is a form of animal behaviour in which the stimulation of a sensory organ (receptor) results in the activity of some organ without the intervention of will.
40. (b) Insectivorous plants mostly thrive in marshes and rocky outcrops or other nitrogen poor areas where sunlight and water are abundant. These plants have evolved the ability to trap and digest insects, which are an excellent source of nitrogen as they contain around 10% nitrogen by mass. Insectivorous plants are able to obtain between 1080% of their total nitrogen from insects.
PART 2 1. Given in ∆ABC, D and E are points on AB , AC respectively such that DE is parallel to BC. BE and CD intersect at O. Area of ∆ADE = 3 sq units Area of ∆DOE = 1sq units Area of ∆BEC = Area of ∆BDC A
3 D
E
1 x
O
x
y B
C
[same base between same parallels] ⇒ (BOD ) + ar (DOE ) = ar (COE ) + ar (DOE ) ⇒ ar (BOD ) = ar (COE ) = x ⇒ ar (BOC ) = y We know that area of two triangle having equal altitude is the same as the ratio of their respective bases. x BO y = = ∴ 1 OE x ⇒ y = x2 Now, ∆ADE and ∆ABC are similar. ar (ADE ) DE 2 ar (ODE ) = = ∴ ar (ABC ) BC 2 ar (OBC ) 3 1 = ⇒ 4 + 2x + y y
KVPY Question Paper 2010 Stream : SA
148 ⇒ 3 y = 4 + 2x + y ⇒ 2 y = 2(2 + x) ⇒ y=x+2 [Q y = x2] ⇒ x2 − x − 2 = 0 ⇒ (x − 2)(x + 1) = 0 ⇒ x = 2, x ≠ −1 ∴Area of ∆ABC = 4 + 2x + y = 4 + 4 + 4 = 12 sq units
2. Let the total number of CD’s sold by the Leela and Madan together = x Total money obtained by them = (x × x ) = x 2 2 They divided x in such that, x2 = 10 (an odd number) + a number less than 10 ⇒ x = 10q + r [Q0 ≤ r < 10] ⇒ x2 = (10q + r )2 ⇒ x2 = 100q2 + 20qr + r 2 2 r = 10 (an odd number ) + a number less than 10 0 ≤ r < 10 ∴Taking r = 0, 1, 2, 3, ..., 9, we get r = 4 or 6 r = 16 or 36 r 2 = 10 + 6 or 3(10) + 6 Hence, the amount left for Madan at the end is 6 rupees.
3. (a) We have, A natural number ‘m’ whose all digits are 1 ∴ m = 1, 111, 1111, 11111, ..., (1111 ... 1) A natural number ‘n’ which is relatively prime to 10 when n + 1number 1, 111, 1111, 11111, …, (1111…1) divides by n, then n + 1remainder obtained. ∴ Possibilities of remainder are 0, 1, 2, 3, ...., n − 1which are n in numbers, where two remainder are same. Let two numbers x =11 1 ... 1 and y = 111 ... 1 having say i digits and j digits respectively which leave the same remainder after division by n. We take, i< j y − x is divisible by n; But y − x = 11... 1000 ... 0 where j − i number of 1’s and remaining zero. Since, n is coprime to 10. We see that n divides m = 111 ... 1 a number having 1’s as its digits. (b) Let the positive rational number is p , where q ≠ 0 q and q = 2r ⋅ 5s ⋅ t where t is coprime to 10 Choose any number m having only is as its digit and is divisible by t.
Consider 9m, which has only 9 as its digits and b divisible by t. 9m Let k= t ∴ 9k = 9m ⋅ 2r ⋅ 5s = (10c − 1)2r ⋅ 5s where, c is the number of digits in m. We can find d such that qd = 10b (10c − 1) [such that 10b is a suitable power of 2 if s > r and a suitable power of s if r > s] p pd a = = ∴ q qd 10b (10c − 1)
So, error in case Q is ⇒ δRQ = Rest − R = RA
where, pd = a.
(c) If R =
(b) In case Q,
I E
V = I (R + RA ) V R = − RA = Rest − RA I
⇒
I
IV
V
5. Given situation is
RV E + –
Current through cell is I = IR + IV = V / R + V / RV where, V = potential drop across resistance R. 1 R + RV 1 I =V + ⇒ =V R RV RRV V RRV = (R + RV ) ⇒ I VRV V ⇒ RV − R = I I VRV ⇒ R= I (RV − V /I ) ⇒
But ∴
RA RV then, 2 δRP Rest R2 = = est ≈1 δRQ RA RV R2
R RA A
V
+ –
4. (a) In case P, IR
A
R
V RV R= . I R − V V I V = Rest = estimated resistance I RV R = Rest RV − Rest −1 R = Rest . 1 − est RV
20 cm f=–10 cm
(a) We use lens formula to get position of image u = − 20 cm, f = − 10 cm. 1 1 1 As, − = v u f 1 1 1 −1 1 = + = − ⇒ v f u 10 20 1 − 2 − 1 −3 = = ⇒ v 20 20 −20 cm ⇒ v= 3 Image is virtual and it is in front of lens 20 at cm. 3 (b) Concave mirror again converge rays and its image formed is coincident with object.
O I
R = Rest 1 + est RV Using binomial approximation and neglecting higher order terms. Now, error in case P is δRP = Rest − R =
2 Rest RV
x
Let mirror is placed at distance x behind the lens. Now, first image I1 of lens acts like object for mirror. Hence, for mirror, 20 u = − + x cm. 3
149
KVPY Question Paper 2010 Stream : SA Given, f = − 5 cm By lens equation, we get position of image I 2 from concave mirror as 1 1 1 + = v u f 1 1 1 or =− + v u f 1 −1 1 ⇒ + = (− 5) v − 20 + x 3 ⇒
20 5 − + x 3 1 1 1 − = = 20 v 20 + x 5 5 + x 3 3 − (3x + 5) − (3x + 5) 1 = = v 15 20 + x 5 (20 + 3x) 3
As this image is acting like an object for lens, so we have from lens equation, 5 (20 + 3x) cm, f = − 10 cm u=− (3x + 5) v = + 20 cm 1 1 1 ⇒ + = v u f (3x + 5) −1 1 − = ⇒ 20 5 (20 + 3x) 10 5 (20 + 3x) − 20 (3x + 5) 1 =− ⇒ 20 × 5 (20 + 3x) 10 100 + 15x − 60x − 100 1 =− ⇒ 2000 + 300x 10
and
2000 = 150 x 2000 =x 150 40 cm = 13.33 cm x= ⇒ 5 So, mirror is placed at 13.33 cm behind the concave lens. (c) When a plane mirror is placed at same position, then For plane mirror is f = ∞, 20 u = − + x 3 20 40 = − + 3 5 60 =− = − 12 cm 5 Now, by mirror equation, 1 1 1 + = v u f 1 1 or =− v u −1 1 = − ⇒ 12 v So,
v = 12 cm
As there is lens in between, so final image formed is calculated as 1 1 1 − = v u f 1 1 1 2 = + =− v −10 12 120 ∴ v = − 60 cm So, final image is formed at a distance of 60 cm left of lens.
6. (a) Mass m is at height H from point Q, where potential energy is taken zero. A θ R
R–h
m h
B Q
From geometry of above figure, if at some angle θ, height of mass m above lowest point Q is h, then from ∆ABC, R−h cosθ = ⇒ h = R (1 − cosθ) R Hence, potential energy of m as a function of θ is PE = U (θ) = mgh ⇒ U (θ) = mgR (1 − cosθ) (b) Kinetic energy at position θ = Loss of potential energy that occurred in reaching this position ⇒ Kinetic energy is K (θ) = mgH − U (θ) ⇒ K (θ) = mgH − mgR (1 − cosθ) = mg (H − R (1 − cosθ)) (c) For H < < R, mass m will oscillate about mean position in SHM. Time period of oscillation of m is 1 R T = 2π g Time taken to travel from P to Q is onefourth of this time period. 1 R i.e. t= 8π g (d) From energy conservation at lowest point, if m has velocity v, then 1 mv2 = mgH 2 mv2 2mgH = ⇒ mv2 = 2mgH or R R But this centripetal force is resultant of force of normal reaction N and weight of body. mv2 N − mg = ⇒ R
⇒ ⇒
2mgH R 2H N = mg 1 + R
N = mg +
This is the force by block on the concave surface.
7. The balanced chemical equations for the given equations can be written as (i) 3Cu + 8HNO3 → 3Cu(NO3 )2 + 2NO + 4H2O (ii) 2CuI2 → Cu 2I2 + I2 (iii) 2Na 2S2O3 + I2 → Na 2S4 O6 + 2NaI Thus, (a) The coefficients are : a = 3, b = 8, c = 3, d = 2 and e = 4 (b) The coefficients are : f = 2, g = 1, h = 1 (c) The coefficients are : i = 2, j = 1, k = 1 and l = 2 2.54 (d) No. of moles of I2 = = 0.01 mol 254 Moles of CuI2 = 2 moles of I2 = 1 mole of Cu = 2 × 0.01 = 0.02 Weight = mole × atomic weight Weight of Cu = 0.02 × 63.5 = 127 . g % of copper present in the alloy . 127 = × 100 = 63.5% 2
8. Conclusions that can be made from the given observations are as follows (a) (i) As the compound in the bottle A did not dissolve in either 1 N NaOH or 1 N HCl. This indicates that the molecule is neutral. Thus, the compound A is (III). CH3
(ii) As the compound in the bottle B dissolved in 1 N NaOH but not in 1 N HCl shows that the compound is acidic in nature, thus the compound in bottle B is (II). CO2H
(iii) The compound in bottle C dissolved in both 1 N NaOH and 1 N HCl indicates that the compound is amphoteric in nature. Thus, the compound in bottle C is (IV). CO2H NH2
(amphoteric because of its Zwitter ion)
KVPY Question Paper 2010 Stream : SA
150 (iv) The compound in bottle D dissolved in both 1 N NaOH, but dissolve in 1 N HCl shows that the compound is a base, thus the compound is (I). NH2
(b) The compound with highest solubility in distilled water is (iv), COOH
= 342 × 1866 . g = 638172 . 1 mole of C12H22O11 gives 12 moles of CO2 ∴1.866 moles of C12H22O11 gives = 1866 . × 12 moles of CO2 = 22.392 moles of CO2 At STP 1 mole of CO2 = 22.4 L
In experiment C, since the lime water turns milky, ethanol fermentation is occurring. In addition, since removal of air did not affect the reaction, the fermentation is anaerobic and yeast must be the organism in the flask. (b) In RBCs, lactic acid fermentation occurs.
22.392 moles of CO2 = 22.4 × 22.392 L
12. (a) The result of radiocarbon dating
L = 5015808 .
10. (a) Difference in flower colour is most NH2
This is due to Zwitter ion formation which can be shown as –
COO +
NH3
9. C12H22O11 (s) + 12 O2 ( g ) → 12CO2 ( g ) + 11H2O(l) ∆H = − 5.6 × 105 J (a) 1 kcal = 418 . kJ ∴2500 kcal = 2500 × 418 . kJ = 10450 kJ So, the energy requirement of the human body per day is 10450 kJ. (b) No. of moles of sucrose 10450 × 103 J mol = 1866 . = 5.6 × 106 J Weight of sucrose = molar mass × number of moles
likely due to environmental factors. (b) Perform crossbreeding between the plants from Chandigarh and those from Shimla to find out whether we get any pink flowers or flowers with any shade of colour between pink and white in the F1 generation. (c) Grow the plants from Chandigarh in Shimla and check whether they still produce white flowers or bear pink flowers.
11. (a) In experiment A, ethanol fermentation occurs producing CO2, turning lime water milky. Since acid is not produced, the dye colour does not change. In experiment B, lactic acid fermentation takes place, which produces acid but does not produce CO2. Hence, dye colour changes to yellow but the lime water does not turn milky.
was correct. Reason Vehicles running on the highway beside the house emitted carbon dioxide from the combustion of petrol or diesel, which are fossil fuels. The carbon in this carbon dioxide, coming from living material that has been converted into petroleum millions of years ago, would get assimilated into the tissues of the plant as it uses carbon dioxide from the surrounding atmosphere for photosynthesis. Therefore, tissues of the plant, when used for radiocarbon dating, would show the age of the plant to be many thousands of years old. (b) A simple experiment to test the validity of this explanation would be to collect seeds from the plants and grow them in a plot of land away from the highway or other sources of CO2 coming from the burning of fossil fuels. Radiocarbon dating of plants growing from these seeds should show them as young plants.
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
QUESTION PAPER 2009 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS x+5 1. The real numbers x satisfying > 1 are precisely 1−x those which satisfy (a) x < 1
(b) 0 < x < 1
(c) − 5 < x < 1 (d) − 1 < x < 1
2. Let tn denote the number of integralsided triangles with distinct sides chosen from {1, 2, 3, ...., n}. Then, t20 − t19 equals (a) 81
(b) 153
(c) 163
(d) 173
3. The number of pairs of reals (x, y) such that x = x2 + y2 and y = 2xy is (a) 4
(b) 3
(c) 2
(d) 1
4. How many positive real numbers x satisfy the equation x3 − 3x + 2 = 0 ? (a) 1
(b) 3
5. Let (1 + 2x)
(c) 4
(d) 6
= a 0 + a1x + a 2x + K + a 20x20.Then, 3a 0 + 2a1 + 3a 2 + 2a3 + 3a 4 + 2a5 + K + 2a19 + 3a 20 equals 20
2
5 ⋅ 320 − 3 5 ⋅ 320 + 3 5 ⋅ 320 + 1 5 ⋅ 320 − 1 (b) (c) (d) (a) 2 2 2 2
6. Let P1 , P2 , P3 , P4 , P5 be five equally spaced points on the circumference of a circle of radius 1, centred at O. Let R be the set of points in the plane of the circle that are closer to O than any of P1 , P2 , P3 , P4 , P5 . Then, R is a (a) circular region (b) pentagonal region (c) rectangular region (d) oval region that is not circular
7. A company situated at (2, 0) in the XY plane charges ` 2 per km for delivery. A second company at (0, 3) charges ` 3 per km for delivery. The region of the plane where it is cheaper to use the first company is (a) the inside of the circle (x + 5.4)2 + y2 = 18.72 (b) the outside of the circle (x + 1.6)2 + ( y − 5.4)2 = 18.72 (c) the inside of the circle (x − 1.6)2 + ( y + 5.4)2 = 18.72 (d) the outside of the circle (x − 5.4)2 + ( y + 1.6)2 = 18.72
8. In a right ∆ ABC, the incircle touches the hypotenuse AC at D. If AD = 10 and DC = 3, the inradius of ABC is (a) 5
(b) 4
(c) 3
(d) 2
KVPY Question Paper 2009 Stream : SA
152 9. The sides of a quadrilateral are all positive integers and three of them are 5, 10, 20. How many possible value are there for the fourth side? (a) 29
(b) 31
(c) 32
(d) 34
10. If the volume of a sphere increases by 72.8%, then its
©
surface area increases by (a) 20%
(b) 44%
(c) 24.3%
(d) 48.6%
11. If the decimal 0. d25d25d25... is expressible in the form n / 27, then d + n must be
(a) 9
(b) 28
(c) 30
(d) 34
12. At what time between 10 O’clock and 11 O’clock are the two hands of a clock symmetric with respect to the vertical line (give the answer to the nearest second)? (a) 10h 9m 13s (c) 10h 9m 22s
(b) 10h 9m 14s (d) 10h 9m 50s
(d)
17. Consider two spherical planets of same average density. Second planet is 8 times as massive as first planet. The ratio of the acceleration due to gravity of the second planet to that of the first planet is (a) 1
(b) 2
(c) 4
18. Two immiscible liquids A and B are kept in an Utube. If the density of liquid A is smaller than the density of liquid B, then the equilibrium situation is
13. A woman has 10 keys out of which only one opens a lock. She tries the keys one after the another (keeping aside the failed ones) till she succeeds in opening the lock. What is the chance that it is the seventh key that works? (a)
7 10
(b)
1 2
(c)
3 10
(d)
(a)
(b)
A
(c)
(d) None of these
students like football and 82% like tennis. Then, all the three sports are liked by at least (a) 68%
(b) 32%
(c) 77%
(d) 36%
15. Let Sn be the sum of all integers k such that 2n < k < 2n only if
+1
(a) n is odd (c) n is even
A
A
1 10
14. In a certain school, 74% students like cricket, 76%
(d) 8
, for n ≥ 1. Then, 9 divides Sn if and (b) n is of the form 3k + 1 (d) n is of the form 3k + 2
19. In the figure given below, a ray of light travelling in a medium of refractive index µ passes through two different connected rectangular blocks of refractive indices µ1 and µ 2 (µ 2 > µ1 ). θ1
PHYSICS 16. A boy standing on the footpath tosses a ball straight up and catches it. The driver of a car passing by moving with uniform velocity sees this.
µ µ1
µ2
θ2
The angle of incidence θ1 is increased slightly. Then, the angle θ 2 is (a) increases (b) decreases (c) remains same (d) increases or decreases depending on the value of (µ 1 / µ 2 )
The trajectory of the ball as seen by the driver will be
20. Two charges of same magnitude move in two circles of radii R1 = R and R2 = 2R in a region of constant uniform magnetic field B0. The work W1 and W 2 done by the magnetic field in the two cases respectively, are such that
(a)
(b)
(a) W1 = W2 = 0 (c) W1 = W2
(b) W1 = W2 ≠ 0 (d) W1 < W2
153
KVPY Question Paper 2009 Stream : SA 21. Two charges + q and − q are placed at a distance b apart as shown in the figure given below. B P
A
The bulb will light up, if (a) (b) (c) (d)
S1 , S2 and S3 are all closed S1 is closed but S2 and S3 are open S2 and S3 are closed but S1 is open S1 and S3 are closed but S2 is open
29. Two bulbs, one of 200 W and the other of 100 W are
C
connected in series with a 100 V battery which has no internal resistance.
b/2
100V +q
–q b
The electric field at a point P on the perpendicular bisector as shown is (a) along vector A (c) along vector B
(b) along vector C (d) zero
22. A block of mass M is at rest on a plane surface
inclined at an angle θ to the horizontal. The magnitude of force exerted by the plane on the block is
(a) Mg cos θ (b) Mg tan θ (c) Mg sin θ (d) Mg
23. We are able to squeeze snow and make balls out of it because of (a) anomalous behaviour of water (b) large latent heat of ice (c) large specific heat of water (d) low melting point of ice
demonstrated by light, but not with sound waves in an air column? (b) Diffraction (d) Polarisation
25. The temperature of a metal coin is increased by 100°C and its diameter increases by 0.15%. Its area increases by nearly (a) 0.15%
(b) 0.30%
(c) 0.60%
same pitch. The property of sound that is most important in distinguishing between the two instruments is (a) fundamental frequency (b) displacement amplitude (c) intensity (d) waveform 235 92 U atom 9
disintegrates to 207 82 Pb with a halflife of 10 yr. In the process, it emits 7 α particles and n β− particles. Here, n is (a) 7
(b) 3
(c) 4
(d) 14
28. Consider the following circuit given below. S1
(a) the current passing through the 200 W bulb is more than that through the 100 W bulb (b) the power dissipation in the 200 W bulb is more than that in the 100 W bulb (c) the voltage drop across the 200 W bulb is more than that across the 100 W bulb (d) the power dissipation in the 100 W bulb is more than that in the 200 W bulb
and equal masses are heated to the same temperature and left to cool in the same surroundings. Then, (a) the cube will cool faster because of its sharp edges (b) the cube will cool faster because it has a larger surface area (c) the sphere will cool faster because it is smooth (d) the sphere will cool faster because it has a larger surface area
(d) 0.0225%
26. The note “Saa” on the Sarod and the Sitar have the
27.
100 W
30. A solid cube and a solid sphere of identical material
24. Which of the following phenomena can be (a) Reflection (c) Refraction
200 W
Then,
CHEMISTRY 31. The element X which forms a stable product of the type XCl4 is (a) Al
(b) Na
(c) Ca
(d) Si
32. A mixture of NH4Cl and NaCl can be separated by (a) filtration (c) sublimation
(b) distillation (d) decantation
33. The pair in which the first compound is ionic and the second compound is covalent, is (a) Fe(OH)2 , CH3 OH (c) CH3 OH, CH3 CH2OH
(b) Fe(OH)2 , Cu(OH)2 (d) Ca(OH)2 ,Cu(OH)2
34. In the reaction, SO2 + 2H2S → 3S + 2H2O, the substance that is oxidised is S2
(a) SO2
(b) H2O
(c) S
(d) H2S
35. Sodium oxide dissolves in water to give sodium S3
hydroxide which indicates its (a) acidic character (c) amphoteric character
(b) basic character (d) ionic character
KVPY Question Paper 2009 Stream : SA
154 36. For an ideal gas, Boyle’s law is best described by (a)
(b) p
p T
(c)
V
(d) p
p V
T
37. The pH values of (i) 0.1 M HCl (aq), (ii) 0.1 M KOH, (iii) tomato juice and (iv) pure water follow the order. (a) (i) < (iii) < (iv) < (ii) (c) (i) < (ii) < (iii) < (iv)
(b) (iii) < (i) < (iv) < (ii) (d) (iv) < (iii) < (ii) < (i)
38. When calcium carbide is added to water, the gas that is evolved is (a) carbon dioxide (c) acetylene
(b) hydrogen (d) methane
39. The atomic radii of the alkali metals follow the order (a) Li > Na > K > Cs (c) Na > K > Cs > Li
(b) K > Cs > Li > Na (d) Cs > K > Na > Li
40. The number of possible structural isomers of C3 H4 is (a) 1
(b) 2
(c) 3
(d) 4
41. Among the four compounds, (i) acetone, (ii) propanol, (iii) methyl acetate and (iv) propionic acid, the two that are isomeric are (a) methyl acetate and acetone (b) methyl acetate and propanol (c) propionic acid and methyl acetate (d) propionic acid and acetone
42. One mole of nitrogen gas on reaction with 3.01 × 1023 molecules of hydrogen gas produces (a) one mole of ammonia (b) 2.0 × 1023 molecules of ammonia (c) 2 moles of ammonia (d) 3.01 × 1023 molecules of ammonia
43. Saponification is (a) hydrolysis of an ester (b) hydrolysis of an amide (c) hydrolysis of an ether (d) hydrolysis of an acid chloride
44. A concentrated solution of lead nitrate in water can be stored in (a) an iron vessel (c) a zinc vessel
45.
(b) a copper vessel (d) a magnesium vessel
Solubility g/L 250 200 150 100
(a) At room temperature, the solubility of KNO3 and KCl are not equal (b) The solubilities of both KNO3 and KCl increase with temperature (c) The solubility of KCl decreases with temperature (d) The solubility of KNO3 increases much more compared to that of KCl with increase in temperature
BIOLOGY 46. Which one of the following is the smallest in size? (a) Bacteria (c) Mammalian cell
(b) Mitochondrion (d) Virus
47. If birds are moved from 30°C  10°C, their body temperature (a) changes from 30°C  10°C (b) increases by 10°C (c) does not change at all (d) decreases by 10°C
48. Ascorbic acid is a/an (a) strong inorganic acid (c) vitamin
(b) hormone (d) enzyme
49. Bile salts (a) breakdown polypeptide chains (b) emulsify fats and solubilise them (c) digest fats (d) help breakdown of polysaccharides
50. Dietary fibres are composed of (a) cellulose (c) amylase
(b) proteins (d) unsaturated fats
51. ‘On the Origin of Species, by Means of Natural Selection’ was written by (a) Hugo de Vries (c) Charles Darwin
(b) Charles Dickens (d) Alfred Russell Wallace
52. Unlike humans, dogs cannot perspire to get rid of excess metabolic heat. They lose metabolic heat by (a) panting (b) taking a bath (c) running in windy conditions (d) rolling in the mud
53. Haemodialysis is a treatment option for patients with malfunctions of (a) kidney (c) heart
(b) liver (d) lungs
54. An individual has ‘O’ blood group if his/her blood sample
KNO3 KCI
50 0
Given the solubility curves of KNO3 and KCl, which of the following statements is not true ?
20 40 60 80 100 Temperature (ºC)
(a) clumps only when antiserum A is added (b) clumps only when antiserum B is added (c) clumps when both antiserum A and antiserum B are added (d) does not clump when either antiserum A or antiserum B is added
155
KVPY Question Paper 2009 Stream : SA 55. In warmer weather, curd from milk forms faster
58. The part of the human brain that governs memory
because
and intelligence is
(a) bacteria diffuse better in warmer milk (b) the rate of bacterial multiplication increases (c) lactogen is better dissolved (d) it is easier to separate protein from water
(a) cerebrum (b) medulla (c) hypothalamus (d) cerebellum
59. Saturated dietary fats increase the risk of heart
56. Seedlings grown in dark are (a) similar to those grown in light (b) taller than those grown in light (c) shorter than those grown in light (d) they do not grow at all
57. In humans, Rhesus condition can arise when (a) father is Rh + and mother is Rh − (b) father is Rh − and mother is Rh + (c) either father or mother is Rh + (d) either father or mother is Rh −
disease by (a) widening arteries by thinning their walls (b) narrowing veins by carbohydrate deposition (c) narrowing arteries by fat deposition (d) narrowing arteries by carbohydrate deposition
60. Rotation of crops is carried out to (a) increase variation in the mineral content of the soil (b) increase diversity of plant habitats (c) increase in nitrogen content of the soil (d) increase convenience for the farmer
PARTII (2 Marks Questions) PHYSICS
MATHEMATICS 61. Let loga b = 4, logc d = 2, where a , b, c, d are natural numbers. Given that b − d = 7, the value of c − a is (a) 1 (c) 2
(b) − 1 (d) − 2
62. Let P (x) = 1 + x + x2 + x3 + x4 + x5 . What is the
66. A spring balance A reads 2 kg with a block of mass m suspended from it. Another balance B reads 3 kg when a beaker with a liquid is put on its pan. The two balances are now so arranged that the hanging mass m is fully immersed inside the liquid in the beaker as shown in the figure given below.
remainder when P (x12 ) is divided by P (x)? (a) 0 (c) 1 + x
(b) 6 (d) 1 + x + x2 + x3 + x4
A
63. In a ∆ ABC, the altitudes from B and C on to the opposite sides are not shorter than their respective opposite sides. Then, one of the angles of ABC is (a) 30° (c) 60°
(b) 45° (d) 72°
B
64. In a ∆ ABC, AB = AC = 37. Let D be a point on BC such that BD = 7, AD = 33. The length of CD is
In this situation,
(a) 7 (c) 40
(a) the balance A will read 2 kg and B will read 5 kg (b) the balance A will read 2 kg and B will read 3 kg (c) the balance A will read less than 2 kg and B will read between 3 kg and 5 kg (d) the balance A will read less than 2 kg and B will read 3 kg
(b) 11 (d) not determinable
65. A line segment l of length a cm is rotated about a vertical line L keeping the line l in one of the following three positions (I) l is parallel to L and is at a distance of r cm from L, (II) l is perpendicular to L and its midpoint is at a distance r cm from L, (III) l and L are in the same plane and l is inclined to L at an angle 30° with its midpoint at a distance r cm from L. Let A1, A2 , A3 be the areas so generated. If r > (a / 2),then (a) A1 < A3 < A2
(b) A2 < A1 < A3
(c) A1 = A3 < A2
(d) A1 = A2 = A3
67. According to the quantum theory, a photon of
electromagnetic radiation of frequency ν has energy E = hν, where h is known as Planck’s constant. According to the theory of relativity, a particle of mass m has equivalent energy E = mc2, where c is speed of light. Thus, a photon can be treated as a hν particle having effective mass m = 2 . c
KVPY Question Paper 2009 Stream : SA
156 If a flash of light is sent horizontally in earth’s gravitational field, then photons while travelling a horizontal distance d would fall through a distance given by (a)
gd 2
mcd 2 (c) h
h (b) mc
2c2
(d) zero
68. A solid square plate is spun around different axes with the same angular speed. In which of the following choice of axis of rotation will the kinetic energy of the plate be the largest? (a) Through the centre, normal to the plate (b) Along one of the diagonals of the plate (c) Along one of the edges of the plate (d) Through one corner normal to the plate
69. An object is placed 0.40 m from one of the two lenses L1 and L2 of focal lengths 0.20 m and 0.10 m respectively, as depicted in the figure. The separation between the lenses is 0.30 m. L2
L1
CHEMISTRY 71. Reaction of NaCl with conc. H2SO4 liberates a gas, X that turns moist blue litmus paper red. When gas X is passed into a test tube containing egg shell powder suspended in water another gas, Y is generated which when passed through lime water makes it milky. The gases X and Y, respectively, are (a) HCl and CO2 (c) SO2 and CO2
(b) Cl 2 and CO2 (d) SO2 and HCl
72. 10 mL of an aqueous solution containing 222 mg of
calcium chloride (mol. wt. = 111) is diluted to 100 mL. The concentration of chloride ion in the resulting solution is
(a) 0.02 mol/L (c) 0.04 mol/L
(b) 0.01 mol/L (d) 2.0 mol/L
73. Aluminium reduces manganese dioxide to manganese at high temperature. The amount of aluminium required to reduce one g mole of manganese dioxide is (a) 1/2 g mol (c) 1 g mol
(b) 3/4 g mol (d) 4/3 g mol
74. Ethanol on reaction with alk. KMnO4 gives X which 0.40m
when reacted with methanol in the presence of an acid gives a sweet smelling compound Y, X and Y respectively, are
0.30m
The final image formed by these two lenses system is at (a) 0.13 m to the right of the second lens (b) 0.05 m to the right of the second lens (c) 0.13 m to the left of the second lens (d) infinity
70. 5 charges each of magnitude 10−5 C and mass 1 kg are placed (fixed) symmetrically about a movable central charge of magnitude 5 × 10−5 C and mass 0.5 kg as shown in the figure given below. The charge at P1 is removed. The acceleration of the central charge is P1
(a) acetaldehyde and acetone (b) acetic acid and methyl acetate (c) formic acid and methyl formate (d) ethylene and ethyl methyl ether
75. The pH of a 10 mL aqueous solution of HCl is 4. The amount of water to be added to this solution in order to change its pH from 4 to 5 is (a) 30 mL
(b) 60 mL
(c) 90 mL
(d) 120 mL
BIOLOGY 76. Proteins are synthesised on (a) cytoskeleton (c) ribosomes
(b) mitochondria (d) Golgi apparatus
77. Which of the following allows light to focus in visual perception? P2
P5 O
(a) Retina (c) Retinal pigment
(b) Iris (d) Cornea
78. During cell division, if there is one round of chromosome P3
[Given,
P4
1 = 9 × 109] OP1 = OP2 = OP3 = OP4 = OP5 = 1 m, 4πε 0 (a) 9 ms −2 upwards (b) 9 ms −2 downwards (c) 4.5 ms −2 upwards (d) 4.5 ms −2 downwards
duplication followed by one round of cell division, the number of chromosomes the daughter cells will have as compared to the mother is (a) equal
(b) double
(c) half
(d) one fourth
79. Similar type of vegetation can be observed, in the same (a) latitude
(b) longitude (c) country
(d) continent
80. Which of the following ecological food chains does not represent an erect pyramid of numbers ? (a) Grass–Rodent–Snake (c) Grass–Deer–Tiger
(b) Tree–Bird–Avian parasite (d) Insect–Chicken–Human
KVPY Question Paper 2009 Stream : SA
Answers PARTI 1
(d)
2
(a)
3
(a)
4
(a)
5
(c)
6
(b)
7
(b)
8
(d)
9
(a)
10
(b)
11
(d)
12
(b)
13
(d)
14
(b)
15
(c)
16
(b)
17
(b)
18
(c)
19
(b)
20
(a)
21
(a)
22
(a)
23
(*)
24
(d)
25
(b)
26
(d)
27
(c)
28
(c)
29
(d)
30
(b)
31
(d)
32
(c)
33
(a)
34
(d)
35
(b)
36
(c)
37
(a)
38
(c)
39
(d)
40
(b)
41
(c)
42
(b)
43
(a)
44
(b)
45
(c)
46
(d)
47
(c)
48
(c)
49
(b)
50
(a)
51
(c)
52
(a)
53
(a)
54
(d)
55
(b)
56
(b)
57
(c)
58
(a)
59
(c)
60
(a)
PARTII 61
(a)
62
(b)
63
(b)
64
(c)
65
(d)
66
(c)
67
(a)
68
(d)
69
(d)
70
(c)
71
(a)
72
(c)
73
(d)
74
(b)
75
(c)
76
(c)
77
(d)
78
(a)
79
(a)
80
(b)
* No option is correct.
Solutions 1. (d) We have, x+ 5 1− x ∴ Again,
>1
x + 5 > 0, 1 − x > 0 x> − 5 x1 1− x
…(i) …(ii)
x + 5 > 1− x x + 5 > (1 − x)2 [squaring both sides] ⇒ x + 5 > 1 + x2 − 2x 2 ⇒ x − 3x − 4 < 0 ⇒ (x − 4) (x + 1) < 0 …(iii) ⇒ x∈ (− 1, 4) From Eqs. (i), (ii) and (iii), we get x ∈ (− 1, 1) ∴ − 1< x < 1 ⇒ ⇒
2. (a) tn denotes the number of integral sided triangle with distincts sides from {1, 2, 3, ..., n}. t19 is the number of triangle formed by the sides from {1, 2, 3, ..., 19} and t20 is the number of triangle formed by the distinct sides from {1, 2, 3, ..., 20}. Any triangle counted in t19 is also counted in t20 , but t20 − t19 is the number of triangle counted in t20 but not in t19 . A triangle is counted in t20 but no t19 if and only if its largest side is 20. The middle side of is a and the smallest side can be 21 − a to a − 1.
So, the number of triangle with largest side 20 and middle side. a = 11, then other sides are 21 − 11, 11 − 1 i.e. 10, 10 (11, 10, 10) 1 triangle. Similarly a = 12, (smallest sides are (9, 10, 11) = 3 triangle a = 13, smallest sides are (8, 9, 10, 11, 12) = 5 triangle ∴Total number of triangles on 1 + 3 + 5 + 7 + .... + 17 = 81
3. (a) We have, x = x2 + y2 and y = 2xy Q y − 2xy = 0 ⇒ y (1 − 2x) = 0 1 y = 0, x = ⇒ 2 Put y = 0 in Eq. (i), we get x = x2 + 0 ⇒ x − x2 = 0 ⇒ x (1 − x) = 0 ⇒ x = 0, x = 1 1 Put x = in Eq. (i), we get 2 2
1 1 = + y2 2 2 1 1 ⇒ y2 = − 2 4 1 1 2 y = ⇒ y=± ⇒ 4 2 ∴ Value of (x, y) are (0, 0) (0, 1) 1 , 1 1 − 1 . 2 2 2, 2
…(i) …(ii)
4. (a) We have, x3 − 3x + 2 = 0 Case I x > 0 ∴ x3 − 3x + 2 = 0 ⇒ (x − 1) (x − 1) (x + 2) = 0 ⇒ x = 1, − 2 Since, x> 0 ∴ x≠−2 x=1 Case II x < 0 ∴ x3 + 3x + 2 = 0 3 Graph of x + 3x + 2 Y
X′
–1
O
X
Y′
Clearly, from graph. It has one solution lie between (− 1, 0). ∴ Positive value of x = 1 Hence, only one solutions.
5. (c) We have, (1 + 2x)20 = a0 + a1 x + a2x2 + ... + a20 x20 Put x = 1, …(i) 320 = a0 + a1 + a2 + ... + a20 Put x = − 1, 1 = a0 − a1 + a2 − a3 + ... + a20 …(ii)
158
KVPY Question Paper 2009 Stream : SA
On adding Eqs. (i) and (ii), we get 320 + 1 = a0 + a2 + a4 + K + a20 2 On subtracting Eq. (ii) from Eq. (i), we get 320 − 1 = a1 + a3 + a5 + K + a19 2 Now, we have 3a0 + 2a1 + 3a2 + 2a3 + K + 2a19 + 3a20 = 3 (a0 + a2 + a4 + K + a20 ) + 2 (a1 + a3 + ... + a19 ) 320 + 1 320 − 1 + 2 = 3 2 2 =
5⋅ 3
20
+ 1
8. (d) We have, ABC is a right angled triangle. AC is hypotenuse of ∆ ABC. The incircle touch the hypotenuse at D. A 10 10 D
C
Given,
P2 Q3
Q2
Q4 O R Q1 Q5
P4
P1
P5
Let R which is near to point O. ∴ OR is lie between the pentagonal region Q1 , Q2 , Q3 , Q4 , Q5 .
7. (b) Let P (x, y) be any point lie in XY plane. P (x, y)
B (0, 3)
P (2, 0)
According to problem, 2PA < 3PB 4PA < 9PB 2
⇒
2
4 [(x − 2)2 + ( y − 0)2 ] < 9 [(x − 0)2 + ( y − 3)2 ]
⇒
4 (x − 4x + 4 + y ) < 9 2
2
(x2 + y2 − 6x + 9)
F r
E
3
r
B
AD = 10 CD = 3 OE = OF = r (radius of circle)
6. (b) P1 , P2 , P3 , P4 , P5 be five equally
P3
r
r
2
spaced points on the circumference of circle of radius 1.
O
3
OE = BE = BF = r AD = AF = 10 [AD and AF are tangents on a circle from external points are equal] Similarly, CD = CE = 3 In ∆ABC, AC 2 = AB 2 + BC 2 (13)2 = (10 + r )2 + (3 + r )2 ⇒ 169 = 100 + 20r + r 2 + 9 + 6r + r 2 ⇒ 2r 2 + 26r − 60 = 0 ⇒ r 2 + 13r − 30 = 0 ⇒ (r + 15) (r − 2) = 0 ⇒ r = 2, r ≠ − 15
9. (a) We have three sides of quadrilateral are 5, 10, 20. Let the fourth sides of quadrilateral = x We know that in quadrilateral. Sum of three side is greater than fourth sides …(i) ∴ 5 + 10 + 20 > x …(ii) 5 + 10 + x > 20 …(iii) 5 + 20 + x > 10 …(iv) 10 + 20 + x > 5 From Eq. (i) x < 35 From Eq. (ii) x > 5 From Eq. (iii) x > − 15 From Eq. (iv) x > − 25 Now, x is a positive integer. ∴ From Eqs. (i), (ii), (iii) and (iv), 5 < x < 35 ∴ Value of x is 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34
⇒
5x2 + 5 y2 − 54 y + 16x + 65 > 0
∴There are 29 possible values of x.
⇒
x + y − 10.8 y + 3.2x + 13 > 0
10. (b) Let initial volume of sphere is V
2
Volume of sphere after increase is V ′ and radius is r′. 4 ∴ V ′ = πr ′3 3 ⇒ V ′ = (V + 72.8% of V ) 728 1728 v = V 1 + = 1000 1000 1728 4 3 V′= ⇒ πr 1000 3 1728 4 3 4 3 ∴ . πr = πr ′ 1000 3 3 r ′3 r′ = 1.728 ⇒ = 1.2 ⇒ 3 r r ∴Increase in surface area of sphere 2 r′ = = 1.44 = 144% r ∴Surface area increase = (144 − 100)% = 44%
11. (d) Let
x = 0. d 25d 25d 25
…(ii) 1000x = d 25. d 25d 25 On subtracting Eq. (i) from Eq. (ii), we get d 25 999x = d 25 ⇒ x = 999 n But given, x= 27 n d 25 d 25 = ⇒n = ∴ 27 999 37 d ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9} n is integer. ∴d25 is a multiple of 37. When, put d = 9 Then, n is multiple of 37. 925 n= = 25 ∴ 37 ∴ n + d = 25 + 9 = 34
12. (b) Exactly at 10 O’clock the hour hand has travelled 300° from 12 O’clock. One hour = 60 minute. One minute hand moves 1° and hour 30 ° 1 ° clock hand move = 12 360 Assuming we have made it to 10 O’clock and now the hour and the minute hand start moving spontaneously. 11
12 1 2
10
3
9
2
⇒ (x + 1.6)2 + ( y − 5.4)2 > 18.72 Hence, the region is outside the (x + 16 . )2 + ( y − 5 ⋅ 4)2 = 18.72
8
4
and radius is r. ∴
V =
4 3 πr 3
…(i)
7
6
5
159
KVPY Question Paper 2009 Stream : SA If the hands of the watch are symmetric with vertical line. Supposing this happens when x minutes have passed x minutes = (6x)° have been covered our hour hand would cover. 1 ° 1 ° (6x) = x 12 2 x ° ∴ Our hand has covered 300 + 2
(2n − 1)(2n ) ⋅ 3 2 But Sn = 9m, m∈ I (2n − 1)2n ⋅ 3 ∴ = 9m 2 ⇒ (2n − 1)2n − 1 = 3m ⇒ 2n (2n − 1) = 6m It is possible when, n is even.
On subtracting this from 360° to find the angle from 12 O’clock anticlockwise, we get x ° x ° 360° − 300 + = 60 − 2 2
16. (b) Velocity of a ball is measured relative to a fixed object or frame. If frame of reference is moving, then object will have a velocity opposite to that of frame of reference. In given case, car is frame of reference. Due to motion of car, the ball has two velocities (i) vertical velocity (ii) horizontal velocity opposite to motion of car. Because of there two velocities, path of ball will be parabolic in car frame.
So, they are symmetric. x 60 − = 6x ∴ 2 120 ° ⇒ x = = 9 min 13.8 s 13 ∴ Time = 10 h 9 m 14 s
length column of B must appear above this line in other limb of Utube. It is as shown below.
Sn =
A
B
This is shown in option (c).
19. (b) Given situation is θ1 A r1
r2
B r3 r4
C
θ2
13. (d) Woman has 10 keys out of which only one opens a lock. 1 10 The conditional probability that the second 1 keys works given that first failed = . 9 ∴ Required probability (seventh key works) = P (I fails) ⋅ P (2nd/I fails) ⋅ P (III /2 fails) P (7th/6th fails) 9 8 7 6 5 4 1 1 = × × × × × × = 10 9 8 7 6 5 4 10 The first keys works with probability
14. (b) Given, 74% students like cricket 76% students like football 82% students like tennis ∴ 26% student not like cricket 24% student not like football 18% student not like tennis Student all the three sports like at least = 100% − (sport not likes) = 100% − (26 + 24 + 18)% = 100% − 68% = 32%
Trajectory of ball is a parabola from front to back
= 8 × mass of first planet … (i) M2 = 8 M1 4 3 4 3 πR2 × ρ = 8 × πR1 × ρ ⇒ 3 3 ∴Density of both planets is same. ⇒ R32 = 8R13 or … (ii) R2 = 2R1 So, ratio of acceleration due to gravity of the second planet to that of the first planet is ⇒
GM2 2 2 g2 R2 M2 R1 = = × g1 GM1 M1 R2 R2 1
− 2 −1
n+1
Last term = 2
−1
2n + 1 − 2n − 1 n ∴ Sn = [2 + 1 + 2n + 1 − 1] 2 2n + 1 − 2n − 1 n (2 )(1 + 2) Sn = 2
When θ1 is increased at point A following Snell’s law (i ∝ r ), then r1 increases. When r1 increases, r2 decreases at point B. Decrease of r2 causes a decrease of r3 as (i ∝ r). Also r3 = r4 , so when r3 decreases, r4 also decreases. This causes a decrease in value of θ2. So, a increase of θ1 causes a decrease in θ2.
20. (a) Force on a charged particle moving in region of magnetic field is F = q(v × B) Clearly, F is perpendicular to both v and B. As, magnetic force is perpendicular to velocity of moving charged particle. So, work done in time ∆t is W = (F ⋅ v) ∆t = 0 Hence, W1 = W2 = 0
21. (a) Given situation is E1 sin θ
So,
8M1 R1 2 × = M1 1 2R1
E2 sin θ θ
g2 = 2 g1 .
18. (c) As density of B is more than that of A, a small volume of B weighs equals to a large volume of A. Hence, if we draw a horizontal line from bottom of column of A, then a lesser
E1 θ E1 cos θ + E2 cos θ θ Resultant
2
=
n
First term = 2n + 1
v1=–vcar
17. (b) Given, mass of second planet
Number of integer between 2n and 2n + 1 is i.e k = 2
Car frame
So, trajectory of ball will be as that of option (c).
15. (c) We have, 2n < k < 2n + 1 , k ∈ N n+1
v2
q
E2 θ q
As charges are of equal magnitude, direction of resultant field is along the angle bisector of the angle formed by field vectors.
160
KVPY Question Paper 2009 Stream : SA
22. (a) Weight of mass M can be resolved into components parallel and perpendicular to inclined plane as shown below. N=M
Mg sin θ
θ Mg cos θ Mg
θ
amplitude rises or falls in a given time duration. This is shown by waveform of sound wave. Two waves can have same amplitude and frequency but can have different waveforms. As shown here. A 1
2 t
Force of block on incline is Mg cosθ, so force exerted by block on mass M is Mg cosθ directed perpendicularly up the plane.
27. (c) Given, 7 α particles are emitted
When we squeeze the snow due to increased pressure, melting point of ice (snow) is lowered and it melts. As pressure is removed, water formed is again frozen because melting point is again raised. This process is called regelation of ice. None of the option given is correct.
from
alignment of electric vectors (or plane of oscillation of particles) in a particular direction when an electromagnetic wave (or a mechanical transverse wave) passes through a narrow slit. Sound waves in air are longitudinal waves, so they cannot be polarised. Polarisation is shown only by transverse waves.
⇒
235 92
U.
235 92
U →
⇒
207 78 A
r12
=
2r1 ∆r + ∆r 2 r12
× 100
As ∆r is small, we can neglect ∆r 2. 2r ∆r ≈ 1 2 × 100 r1 ∆r = 2 × 100 r1 = 2 × 015 . % = 0.30%
26. (d) Property of sound that makes difference between two sounds having same frequency and amplitude is called timber or quality of sound. This is how
→
207 82 Pb
+ n (−01 β )
28. (c) In given circuit, when S1 and S3 both are closed, then source is short circuited. Closing either S1 or S3 does not completes any of path. Hence, they produce no change in circuit. S1
product of the type XCl 4 , so X must be tetravalent. Among the given elements Al is trivalent, Na is monovalent, Ca is divalent and Si is tetravalent. Thus, option (d) is correct. 32. (c) A mixture of NH4 Cl and NaCl can be separated by the process of sublimation. In this process, solid directly changes to gaseous state without passing into liquid state. NH4 Cl sublimes to gaseous NH3 and HCl upon heating whereas NaCl does not sublime, the reaction can be written as NH4 Cl(s) → NH3 ( g ) + HCl( g )
33. (a) An ionic compound is formed when metal and a nonmetal reacts whereas covalent compound is formed when 2 nonmetals react with each other. In the given options, Fe(OH)2, Ca(OH)2 and Cu(OH)2 are ionic while CH3 OH, CH3 CH2OH are covalent. Thus, the correct pair in which first compound is ionic and the second compound is covalent is given in option (a).
34. (d) In the given reaction,
S2
Reduction
S3
× 100
where, ∆r = increase in radius. r 2 + 2r1 ∆r + ∆r 2 − r12 × 100 = 1 r12
+ 7(42 He)
Conservation of atomic number, gives 78 = 82 − n ⇒ n = 82 − 78 = 4 So, 4 β − particles are emitted in given nuclear reaction.
Final area − Initial area = × 100 Initial area π (r22 − r12 ) = × 100 πr12 =
207 78 A
Now, nβ − particles are emitted to produce 207 82 Pb.
25. (b) Percentage increase in area
(r1 + ∆r )2 − r12
30. (b) Area of cube is more than that of a sphere for same mass and density. Cube also have sharp edges that radiates more effectively than a flat surface. So, rate of cooling for cube is much rapid than sphere. Effect of sharp edges is prominent only at very high temperatures. So, option (b) is correct. 31. (d) As element X is forming a stable
23. (No option is matching)
24. (d) Polarisation is a process of
So, power dissipation is more in the 100 W bulb. This makes option (d) correct.
SO2 + 2H2S → 3S + 2H2O Oxidation
On closing S2 alone the circuit containing bulb is complete. Hence, bulb will light up when S2 is closed alongwith S3 but S1 is open. Option (c) is correct.
29. (d) Resistance of bulb is inversely proportional to its rated power. ⇒
R=
2 V rated 1 ⇒R ∝ Prated Prated
As bulbs are in series, so same current flows through them at all instances. Power dissipation in a series combination is
or
P = I 2R ⇒ P ∝ R 1 P∝ Prated
Here, H2S is getting oxidised to H2O.
35. (b) Na 2O + H2O → 2NaOH Sodium Oxide
Sodium hydroxide
As Na 2O is a metal oxide which on dissolving with water gives NaOH which is metal hydroxide, indicates the basic character of Na 2O. 36. (c) According to Boyle’s law, the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature. 1 p∝ ∴ V Both option c and b is showing relationship between p and V but the correct option is (c), as it is showing
161
KVPY Question Paper 2009 Stream : SA inverse relationship and option (b) is showing a linear relationship.
have different functional groups, so they are functional isomers.
42. (b) N2 + 3H2
p
V
37. (a) pH stands for potenz (power) of hydrogen. The pH values for acidic solution ranges from 0 to 7, for pure water its value is 7 and for basic solution, it ranges from 714. As HCl is a strong acid, the pH value will be least. On the other hand, KOH is a strong base, its pH value will be maximum, tomato juice contains citric acid which is a weak acid, so its pH value will be more than HCl, but will be less than H2O and KOH. Thus, the correct order of pH values will be (i) < (iii) < (iv) < (ii)
38. (c) When calcium carbide is added to water, acetylene gas is evolved. The reaction can be written as CaC2 + 2H2O → Ca(OH)2 + CH≡≡ CH (Acetylene)
(Calcium carbide)
39. (d) On moving down the group, as the atomic number of alkali metal increases, their atomic radii also increases because here shielding effect predominates over nuclear charge. Thus, the correct order of atomic radii of alkali metal is Cs > K > Na > Li
40. (b) Two structural isomers of C3 H4 are possible, which are propyne and cyclopropane. CH3 C ≡≡ CH (Propyne)
(Cyclopropene)
41. (c) The molecular formulas for the given compounds are as follows acetone – C3 H6 O propanol – C3 H8 O methyl acetate – C3 H6 O2 propionic acid – C3 H6 O2. As molecular formula for both propionic acid (CH3 CH2 COH) and methyl acetate O functional (CH3 COCH3 ) are same but O

2NH3
6.022 × 10 molecules of H2 = 1mole ∴3.011 × 1023 molecules of H2 = 0.5 mole. 3 moles of H2 reacts with 2 moles of NH3 . 2 So, 1 mole of H2 reacts = moles of NH3 . 3 2 ∴0.5 mole of H2 react with × 0.5 3 1 = mole 3 1 mole of NH3 contains 6.022 × 1023 molecules. 1 So, mole of NH3 will have 3 1 = 6.022 × 1023 × 3 = 2.0 × 1023 molecules. 23
43. (a) Saponification is the process of hydrolysis of glyceryl ester of stearic acid with sodium hydroxide, which produces glycerol and soap. The reaction can be written as O CH2—O—C—C17H35 O CH—O—C—C17H35 CH2—O—C—C17H35 O
+3NaOH
∆
3C17H35COO– Na+ (soap) + CH2OH CHOH CH2OH Glycerol
44. (b) A concentrated solution of lead nitrate in water can be stored in copper vessel as copper can not react with lead nitrate solution because it is less reactive than lead. Whereas, all other given metals are more reactive than lead and can easily react (displace) it from its solution. 45. (c) The solubility of KCl increases with increase in temperature. Thus, the statement (c) is incorrect. (a) From the graph, it is clear that at room temperature the solubility of KNO3 and KCl are not equal. Thus, the statement (a) is correct. (b) The solubility of both KNO3 and KCl increases with temperature. Thus, statement (b) is correct. (d) From the graph, it can be seen that the solubility of KNO3 increases
much more compared to that of KCl with increase in temperature. Thus, the statement (a) is correct.
46. (d) Virus is the smallest in size among bacteria, mammalian cell and mitochondrion. Virus ranges in size from about 20400 nm in diameter. Bacterial cells range in size from 0.210 µm. Mammalian cells are between 10100 µm in diameter. Mitochondria are commonly between 0.753 µm in diameter.
47. (c) Birds and mammals are endothermic animals, i.e. their core body temperature is kept nearly constant through thermal homeostasis. Therefore, if birds are moved from 30°10°C their body temperature does not change at all.
48. (c) Ascorbic acid is also known as vitaminC, a vitamin found in citrus fruits and is an essential nutrient involved in the repair of connective tissue and the enzymatic production of certain neurotransmitters. In the body, it acts as an antioxidant, helping to protect cells from the damage caused by free radicals.
49. (b) Bile salts consist of sodium bicarbonate, sodium glycocholate and sodium taurocholate. These are the primary components of bile produced in the liver. Their main function is emulsification of fat (i.e. breakdown of large fat molecules into small droplets) and their solubilisation. 50. (a) Dietary fibres are composed of nonstarch polysaccharides such as cellulose, dextrins, inulin, lignin, chitins, pectins, betaglucans, waxes and oligosaccharides. Dietary fibres, also known as roughage is the portion of plant derived food that cannot be completely broken down by digestive enzymes. Rest others, i.e. amylase, proteins and unsaturated fats are although basic components of diet but are not considered as dietary fibres.
51. (c) ‘On the Origin of Species, by Means of Natural Selection’ or ‘The Preservation of Favoured Races in the Struggle for Life’, published on 24 November 1859 is a work of scientific literature by Charles Darwin which is considered to be the foundation of evolutionary biology.
52. (a) Panting refers to breathing quickly and loudly through mouth. It helps dogs to get rid of excess metabolic heat when they are hot or engaged in vigorous exercise.
162 This happens because panting helps dogs to circulate the necessary air through their bodies to cool down.
53. (a) Haemodialysis is a process of purifying the blood of a person whose kidneys are not working normally. It helps to filter waste, removes extra fluid and balances electrolytes (sodium, potassium, bicarbonate, chloride, calcium, magnesium and phosphate). During haemodialysis, blood is removed from the body and filtered through a manmade membrane called a dialyser, or artificial kidney and then the filtered blood is returned to the body.
54. (d) An individual has ‘O’ blood group if his/her blood sample does not clump when either antiserum ‘A’ or antiserum ‘B’ is added. Antiserum is a blood serum containing antibodies against specific antigens. Clumping or Agglutination is the process that occurs if an antigen is mixed with its corresponding antibody. A person with ‘O’ blood group neither have ‘A’ nor ‘B’ antigens on his/her red blood cells, but both ‘a’ and ‘b’ antibodies in his/her plasma. Thus, they are also called as universal donor.
55. (b) The rate of bacterial multiplication increases in warmer weather thereby forming curd from milk faster. This happens because Lactobacillus (i.e. bacteria that turns milk into curd) is more active in summers and develops more, which inturn accelerates the fermentation process.
56. (b) Seedlings grown in dark are taller than those grown in light as they develop long hypocotyls (embryonic shoot) and their cotyledons remain closed around the epicotyl in an apical hook. This process is referred to as etiolation.
57. (c) In humans, Rhesus condition can arise when either father or mother is Rh positive. Rhesus (Rh) factor is an inherited protein found on the surface of red blood cells. If an individual’s blood has the protein, he/she is Rh positive and if his/her blood lacks the protein, he/she is Rh negative. A baby may have the blood type and Rhfactor of either parent or a combination of both parents. Rh factor follows a common pattern of law of dominance because Rhpositive gene is a dominant gene.
KVPY Question Paper 2009 Stream : SA 58. (a) The part of the human brain that
63. (b) ABC is a triangle in which BE
governs memory and intelligence is cerebrum. It is the largest part of the human brain associated with most critical and intelligent brain function such as thoughts and actions.
and CF are altitude.
59. (c) Saturated dietary fats increase the risk of heart disease by narrowing arteries by fat deposition. Saturated fat raises the level of cholesterol in your blood. High levels of Low Density Lipid (LDL) cholesterol in blood are responsible for their deposition in blood vessel, thus narrowing the lumens and increasing the risk of heart diseases and stroke. 60. (a) Rotation of crops is carried out to increase variation in the mineral content of the soil. Prolonged planting of the same crop type leads to the depletion of specific nutrients in the soil.
61. (a) We have, log a b = 4, log c d = 2, a , b, c, d ∈ N ⇒
b = a 4 , d = c2
⇒
b − d = 7 = a 4 − c2
⇒
7 = (a 2 + c) (a 2 − c)
⇒
7 × 1 = (a 2 + c) (a 2 − c)
∴ a 2 + c = 7 and a 2 − c = 1 On solving, we get a = 2 and c = 3 ∴ c− a= 3− 2=1
62. (b) We have, P (x) = 1 + x + x2 + x3 + x4 + x5 P (x ) =
1 − x6 1− x
A
B
C
BE ≥ AC CF ≥ AB BE In ∆ABE, sin A = AB AB sin A = BE AB sin A ≥ AC [Q BE ≥ AC] …(i) Similarly in ∆ACF, CF sin A = AC ⇒ AC sin A = CF ⇒ AC sin A ≥ AB [QCF ≥ AB] …(ii) From Eqs. (i) and (ii), we get (AB + AC ) sin A ≥ (AB + AC ) ⇒ sin A ≥ 1 ⇒ sin A = 1 [0 ≤ sin A ≤ 1] ⇒ A = 90° [Q sin 90° = 1] Now, from Eqs. (i) and (ii), we get AB ≥ AC and AC ≥ AB ∴ AB = AC Hence, angles are 45°, 45°, 90°. Given,
64. (c) Given, AB = AC = 37 AD = 33 BD = 7
a ( − rn) 2 n Q a + ar + ar + K + ar = 1− r
A
It has 5 roots let α1 , α 2 , α3 , α 4 , α5 they are 6th roots of unity Now,
R (α1 ) = 6 = R (α 2 ) = R (α3 ) = R (α 4 ) = R (α5 ) ∴ R (x) − 6 = 0 has 6 roots, which contradicts that R (x) is maximum of degree 4. ∴So it is an identity. Q R (x ) = 6
37
37
P (x12 ) = 1 + x12 + x24 + x36 + x48 + x60 ∴ P (x12 ) = P (x) ⋅ Q (x) + R (x) Here, R (x) is a polynomial of maximum degree 4. Put x = α1 , α 2 , α3 , α 4 , α5 ; we get
E
F
33
B 7
D
E
C
In ∆ABE, AB 2 = AE 2 + BE 2
…(i)
AD 2 = AE 2 + DE 2
…(ii)
In ∆ADE, ⇒
AB − AD 2 = BE 2 − DE 2
⇒
AB 2 − AD 2 = (BE + DE ) (BE − DE )
2
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KVPY Question Paper 2009 Stream : SA ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
AB 2 − AD 2 = (CE + DE )(BD ) [Q BE = CE ] AB 2 − AD 2 = CD ⋅ BD AB 2 − AD 2 CD = BD 372 − 332 [given] CD = 7 (37 + 33) (37 − 33) CD = 7 70 × 4 CD = = 40 7
65. (d) Case I Area is generated by line segment l is curved surface area of cylinder = 2πra ∴ A1 = 2 πra
1 B /C = 2 a /2 9 BC = 4 a a A3 = πa r − + r + 4 4 A3 = 2 πra A1 = A2 = A3
⇒ ∴ ⇒ ∴
66. (c) When block is dipped in water, it displaces some water which exerts buoyant force on block. As a result, reading on scale A will be lower than 2 kg. Due to reaction of block (which is equal to buoyant force), beaker of water is pushed. So, reading of scale B will be more than 3 kg. 67. (a) In time t, a particle of mass m falls by a distance 1 2 gt 2 Now, distance covered horizontally = d and speed in horizontal direction = c. d Time to travel distance, d = t = c d
By lens equation, 1 1 1 − = v u f 1 1 1 1 1 or = + = + v f u 0.2 (−0.4) 1 1 ⇒ = v 0.4 ⇒ v = 0.4 m Now, this image acts like a virtual object for second lens. For second lens, u = + 0.1m, f = − 0.1m
h=
a r L
123
Case II Area is formed in circle.
a/2
h=
2
d 1 2 1 gd 2 gt = × g × = c 2 2 2c2
68. (d) Kinetic energy of a rotating body is L
K =
Area of circular region 2 2 a a A2 = π r + − r − 2 2
1 2 Iω 2
ω
= 2πra Case III Area generated to form a frustrum. r–a/4 a/2 B
a/4 C 30°
r
a/2
A
In ∆ABC, BC sin 30° = AC
u=+0.1 m
h
So, photons falls through a distance, r
Final image at ∞
By lens equation, 1 1 1 1 1 = + = + v f u − 01 . 01 . 1 =0 ⇒ v=∞ ⇒ v Hence, final image is formed at infinity. Alternate Method For lens L1 , u = − (2f ) so image is formed at 2f distance. Now, image distance for L2 is 0.1 m and focal length of L2 is also 0.1 m. Hence, object for L2 is at focus, so its image is formed at infinity.
70. (c) Forces on charge at point O,
where, I = moment of inertia of a body. For square plate, moment of inertia is largest along an axis perpendicular to plane of plate and through its one of corner. As mass distribution is now farthest from axis of rotation. So, kinetic energy of plate is largest in option (d).
69. (d) Image of first lens acts like object for second lens. Now, for first lens, u = − 0.40 m, f = + 0.20 m
initially balances each other as it is given that acceleration occurs when charge at point P1 is removed. This means resultant of force due to charges at points P2 , P3 , P4 and P5 is equal and opposite to force due to at point P1 . Hence, acceleration of charge at point O is directed along OP1 . F Acceleration = m Kq1 q2 2 r 9 × 109 × 10−5 × 5 × 10−5 = = m (1)2 × 1 = 4.5 ms −2
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KVPY Question Paper 2009 Stream : SA
71. (a) When concentrated sulphuric acid reacts with NaCl, then sodium bisulphate and HCl gas (X) is formed. The HCl gas released is acidic in nature and turns blue litmus red. 2NaCl + H2SO4 → Na 2SO4 + 2HCl (X )
When HCl gas (X) is passed into a testtube containing egg shell powder which contains calcium carbonate suspended in water another gas CO2 (Y) is released, which turns lime water milky. HCl + CaCO3 → CaCO2 + CO2 + H2O (X )
(Y )
Egg shell powder
4 moles of Al 3 ⇒ Amount of Al required to reduce 1 g 4 mole of MnO2 = g mol. 3
∴ 1 mole of MnO2 reacts =
74. (b) Ethanol on reaction with alkaline KMnO4 gives acetic acid (X), which when reacts with methanol in the presence of an acid gives methyl acetate (Y), which is a sweet smelling compound. Alkaline
CH3 CH2OH → CH3 COOH(X ) Ethanol
KMnO4
Acetic Acid H+
CH3 COOH + CH3 OH → Methanol
CH3 COOCH3 (Y ) + H2O
CO2 + Ca(OH)2 → CaCO3 + H2O
Methyl acetate (Sweet smell)
Milky
Thus, the gases X and Y respectively are HCl and CO2.
Thus, X and Y respectively are acetic acid and methyl acetate.
72. (c) Initial concentration of CaCl 2 in
75. (c) Given,
solution =
222 × 10−3
111 × 10 × 10−3
= 0.2 M
On dilution, M1V1 = M2V 2 0.2 × 10 = M2 × 100 M2 = 0.02 M ∴ On dilution, the final concentration of CaCl 2 will become 0.02 M. CaCl 2 → Ca 2+ + 2Cl − 0.02M
0.02M
2 × 0.02
−
∴ The concentration of Cl ion in the resulting solution = 0.04 mol/L.
73. (d) 4Al + 3MnO2 → 3Mn + 2Al 2O3 3 moles of MnO2 reacts with 4 moles of Al.
pH of 10 mL of HCl solution = 4 ∴Concentration of H+ ions = 10−4 M After dilution, pH of HCl becomes = 5 ∴Concentration of H+ ions = 10−5 M As we know, M1V1 = M2V 2 = 10−4 × 10 = 10−5 × V 2 V 2 = 100 mL So, 90 mL of water should be added for the pH change from 4 to 5.
76. (c) Ribosomes are the site where RNA is translated into protein. This process is called protein synthesis. Protein is needed for many cell functions such as repairing damage or directing chemical processes.
Ribosomes can be found floating within the cytoplasm or attached to the endoplasmic reticulum.
77. (d) The cornea acts as the eye’s outer most lens. It functions like as window that controls and focusses the entry of light into the eye (visual perception). The cornea contributes between 6575 per cent of the eye’s total focussing power. When light strikes the cornea, it bends or refracts the incoming light on to the lens. The cornea covers the pupil (the opening at the centre of the eye), iris (the coloured part of the eye), and anterior chamber (the fluid filled inside of the eye).
78. (a) One round of chromosome duplication followed by one round of cell division leads to equal number of chromosomes in the daughter cells as compared to the mother cell.
79. (a) Latitude and temperature are related to each other in a way that, as we approaches the equator, the temperature gets warmer and as we approaches the poles, it gets cooler. Since, same vegetation grows in the same climatic zone, therefore similar type of vegetation can be observed in the same latitude.
80. (b) Tree–Bird–Avian parasite does not represent an erect pyramid of number. Instead, it is an inverted pyramid of number. All ecological pyramids of number are erect except in parasitic food chain, where one primary producer supports numerous parasites which further support more hyperparasites.
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KVPY Practice Set 1 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
PRACTICE SET 1 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS
4. When 10 is subtracted from each of the given
1. If x = 3 + 1, then value of x4 + (a) 54 (c) 58
16 4
x
is
(b) 55 (d) 56
2. Three friends Ajay, Vijay and Sanjay move along a circular path of length 1.2 km with speeds of 6 km/h, 8 km/h and 9 km/h respectively. Ajay and Vijay move in the same direction but Sanjay move in opposite direction, if they all start at the same time and from same place. How many time will Ajay and Sanjay meets anywhere on the path by the time Ajay and Vijay for the first time anywhere on the path? (a) 6 times (c)8 times
(a) 25
(b) 30
cube touch by all vertical faces with same bases and height, then the ratio of their volume will be (b) 42 : 33 : 11 (d) None of these
(c) 60
(d) 65
5. In the given figure, AB is the diameter of the circle centered at O. If ∠COA = 60° , AB = 2r , AC = d and CD = l, then l is equal to B
D
O 60°
C
A
(b) 7 times (d) 9 times
3. A cone is within the cylinder and cylinder is within a
(a) 14 : 11 : 13 (c) 56 : 36 : 22
observation, the mean is reduced to 60%. If 5 is added to all the given observation, the mean will be
(a) d 3
(b)
d 3
(c) 3d
(d)
3d 2
6. The sum of all integers x for which x4 + x3 + x2 + x + 1 is a perfect square (a) 2
(b) 4
(c) 3
(d) 6
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KVPY Practice Set 1 Stream : SA
7. The sum of all 3digit numbers which are equal to 11 times the sum of squares of their digits is (a) 1212
(b) 1353
(c) 1452
(d) 1364
8. If 2f (xy) = { f (x)} + { f ( y)} for all x, y ∈ R, and f (1) = 2, y
x
then the value f (5) − f (3) is equal to (a) 12
(b) 24
(c) 36
(d) 48
9. A job has to completed by 12 boys in 15 days. If three boys are absent from the first day, then by what percentage should the remaining boys increase their rate of working to complete the job 1 (a) 33 % 3 2 (c) 40 % 3
PHYSICS 16. A solid cylinder of mass m and length l is placed vertically on the ground.
(a)
m2 g 2l 3AY
(b)
m2 g 2l 6AY
functions (a) 2499 (c) 2501
(b) 2500 (d) None of these
11. In a regular heptagon ABCDEFG the side of heptagon is 1, then diagonals (b)
2 3
1 1 is equal to + AC AD
(c) 1
(d)
1 4
12. In the given figure, the length of AB is
(b) 2 6
m m
This wire passes through slab without splitting it into two pieces. This is due to (a) depression of melting point (b) elevation of melting point (c) high conductivity of metal wire (d) high specific heat of ice
18. Two identical boxes one of them is filled with nitrogen and other is filled with helium are put on a fast moving train. If train is suddenly stopped, then what will be the ratio of rise of temperature of two boxes nearly? (b) 1 : 4
(c) 3 6
(c) 4 : 7
p a
b
(d) 4
V
13. There are 20 units cubes all of whose faces are white, and 44 units cubes all of whose faces are red. They are put together to form a bigger cube (4 × 4 × 4). What is the minimum number of white visible on this larger cube? (b) 14
(c) 12
(d) 8
14. A larger tanker can be filled by two pipes A and B in 60 min and 40 min respectively. How many minutes will take to fill the empty tanker if only B is used in the first half of the time and A and B are both used in the second half of the time? (a) 15
(b) 20
(c) 27.5
Temperature of gas during the above expansion process (a) decreases continuously (b) increases continuously (c) decreases then increases (d) increases then decreases
20. An elastic ball of mass m is suspended with an ideal thread. Another ball of same mass hits ball with velocity v0 as shown below. m
(d) 30
37° v 0
15. If α and β are acute angles such that cos2 α + cos2 β = equals (a)
π 6
(b)
3 1 and sin α − sin β = , then α + β 2 4
π 4
(c)
π 3
(d) 1 : 1
19. A gas expands from state a to state b as shown below.
4 cm
6 cm
(a) 20
m2 g 2l AY
placed over a slab of ice.
(a) 2 : 1
B 3 cm
A
(a) 3
(d)
1 (b) 22 % 2 1 (d) 30 % 3
x x , where x is greatest integer condition = 99 101
1 2
m2 g 2l 2AY
(c)
17. A metallic wire is loaded at ends with two masses is
10. The number of positive integer x which satisfies the
(a)
l
If Y = Young’s modulus of cylinder’s material, then strain energy stored is
(d)
π 2
m
Impulsive tension in the string due to collision is (a) mv0
(b)
5 mv0 7
(c)
18 mv0 17
(d)
4 mv0 9
169
KVPY Practice Set 1 Stream : SA 21. A stationary radioactive nucleus decays as:
27. A body falls through a viscous fluid starting from rest
X → +Y If speed of αparticle is v, then speed of daughter nucleus Y will be
towards ground. Then, after a long time, which of the most likely to be correct?
4 2He
4v (a) A−4 4v (c) A+ 4
(a) No energy is dissipated by body (b) Rate of potential energy dissipation is constant (c) Rate of kinetic energy dissipation is constant (d) Whole of the energy of body can be dissipated before reaching ground
2v (b) A−4 2v (d) A+ 4
22. 30 g ice at 0°C is mixed with 25 g of steam at 100°C. The resulting mixture is (Latent heat of fusion = 80 cal/g, latent heat of vapourisation = 540 cal/g and specific heat of water = 1 cal)
D B A l
If length of each block is l, then maximum possible projection for topmost block is (a)
23. An object falling freely from rest covers a distance s in 5th second. Then, distance travelled by object in 7th second is 9 (b) s 13
shown below. C
(a) water and ice at 0°C (b) water at 100°C (c) water and steam at 100°C (d) water, ice and steam at 50°C
13 (a) s 9
28. Identical blocks of wood are piled over each other as
9 (c) s 11
11 (d) s 9
l 2
(b)
l 3
(c)
l 4
(d)
l 6
29. For a two particle systems, kinetic energy K and potential energy U varies with separation r of particles as
24. A pendulum bob is given a push when it is suspended freely. Energy
K
l m
u
(b) 5 : 1
(c) 7 : 1
B
E
C U
The system is a bound system for
If bob successfully completes the vertical circle, then least ratio of kinetic energies at bottom and top of the circle is (a) 2 : 1
A
(d) 1 : 2
(a) r = rA (c) r = rC
(b) r = rB (d) all points A, B and C
30. A point source of light S, placed at a distance L in front of the centre of a plane mirror of width 1 m, hangs vertically on a wall.
25. A wood block is floating in benzene at 0°C.
d = 1m
A man walks in front of mirror along a line parallel.
It is given, Density of wood = 880 kg m −3 Density of benzene = 900 kg m
−4
K
−1
Cubical expansion coefficient of benzene . × 10−3 K −1 = 12 Minimum temperature at which wooden block just sink in is (b) 10°C
(c) 12°C
(d) 15°C
26. Mass of the largest stone that can be moved by flowing water stream depends on density of water, acceleration due to gravity and velocity of flow. Then, mass m is proportional to (a) v2
(b) v4
(c) v6
S 2L
−3
Cubical expansion coefficient of wood = 15 . × 10
(a) 22°C
L
(d) v−1
To the mirror at a distance 2L from mirror as shown in the above figure. Distance upto which source is visible to man is (a) 1 m
(b) 2 m
(c) 3 m
(d) 4 m
CHEMISTRY 31. The correct order of the lattice energies of the following ionic compounds is (a) NaCl > MgBr2 > CaO > Al 2O3 (b) NaCl > CaO > MgBr2 > Al 2O3 (c) Al 2O3 > MgBr2 > CaO > NaCl (d) Al 2O3 > CaO > MgBr2 > NaCl
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KVPY Practice Set 1 Stream : SA
32. Average volume available to a molecule in a sample
43. Clemmensen reduction of ketone is carried out in the presence of which of the following reagents?
of ideal gas at STP is (a) 3.72 × 10−20 cm3 (c) 22400 cm3
(b) 2. 69 × 1019 cm3 (d) 22400 × 6.02 × 1023 cm3
33. Among the quantities, boiling point (I), entropy (II),
(a) ZnHg with HCl (b) LiAlH4 (c) H2 and Pt as a catalyst (d) Glycol with KOH
44. Standard electrode potential of three metals X, Y and
pH (III) and emf of a cell (IV), intensive properties are
Z are − 1 . 2 V, + 05 . V and − 30 . V, respectively. The reducing power of these metals will be
(a) Both I and II (c) I, III and IV
(a) Y > X > Z (c) X > Y > Z
(b) I, II and III (d) All of these
34. The number of radial nodes of 3s and 2p orbitals are (b) 0, 2
(c) 1, 2
(d) 2, 1
35. A sulphur containing species that cannot be a reducing agent is (a) SO2
45. When white phosphorus is heated with caustic soda, the compounds formed are
respectively (a) 2, 0
(b) Z > X > Y (d) Y > Z > X
(b) SO32−
(c) H2SO4
(d) S2−
36. The energy (in J) corresponding to light of
(a) PH3 + NaH2PO3 (c) PH3 + Na 2HPO3
(b) PH3 + NaH2PO2 (d) PH3 + NaH2PO4
BIOLOGY 46. During urine formation, the filtration of blood at the
wavelength 45 nm, is closest to
glomerulus is
(h = 6.63 × 10−34 Js, speed of light = 3 × 108 ms−1 ) (a) 6.63 × 108 (b) 6.67 × 1011 −15 (c) 4.42 × 10 (d) 4.42 × 10−18
(a) an active process (b) an osmotic process (c) a pressure dependent physical process (d) a nonenergymediated transport process
37. The reaction N2 + 3H2 → 2NH3 is used to produce ammonia. When 450 g of hydrogen was reacted with nitrogen, 1575 g of ammonia were produced. The percentage yield of reaction is closest to
(a) 61.8
(b) 72.4
(c) 51.8
(d) 89.1
38. The number of isomers for the compound with the (b) 4
(c) 5
(d) 6
39. Time required to deposit one millimole of aluminium metal by the passage of 9.65 A through molten electrolyte containing aluminium ion is (a) 30 s
(b) 10 s
(a) insufficiency of thyroid hormones (b) excess of thyroid hormones (c) insufficiency of corticosteroids (d) excess of growth hormones
48. ‘Imperfect fungi’ is a group represented by fungal
molecular formula C2BrClFI is (a) 3
47. Grave’s disease is associated with
(c) 30,000 s
(d) 10,000 s
40. The IUPAC name of the following compound is
species which have (a) simple mycelia (b) no known mechanism of sexual reproduction (c) unknown phylogenetic relationship (d) lost its survival mechanism against harsh environment
49. Which of the following is not a characteristic of phylum Chordata? (a) Pharyngeal slits (c) Postanal tail
(a) 4, 4, 3trimethyl hex1yne (b) 4, 4, 3trimethyl hex1ene (c) 3, 4, 4trimethyl hex1yne (d) 3, 4, 4trimethyl hex1ene
50. The energy rich fuel molecules produced in the TCA cycle are
41. Which of the following compounds will not undergo aldol condensation? (a) Methanal (c) Cyclohexanone
(b) 2methyl pentanal (d) 1phenyl propanone
42. Which is the most basic compound among the given options? NO2 N  H I (a) I
(b) II
(b) Amniotic egg (d) Notochord
(a) 2 GTP, 2 NADH and 1 FADH2 (b) 1 GTP, 2 NADH and 2 FADH2 (c) 1 GTP, 3 NADH and 1 FADH2 (d) 2 GTP and 3 NADH
51. In Drosophila melanogaster males, homologous chromosomes pair and segregate during meiosis but crossing over does not occur. At which stage of meiosis does segregation of 2 alleles of a gene take place in their individuals? (a) Zygotene (c) AnaphaseI
(b) Diakinesis (d) AnaphaseII
52. Excess oxygen consumed after a vigorous exercise is N II
III (c) III
(d) Both I and III
(a) to pump out lactic acid from muscle (b) to increase the concentration of lactic acid in muscle (c) to reduce dissolved carbon dioxide in blood (d) to make ATP for gluconeogenesis
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KVPY Practice Set 1 Stream : SA
(d) Evolution need not always lead to a better phenotype
53. Mark the correct relationship (a) ψw = ψ p − (ψπ + ψm ) (c) ψw = ψ p + ψπ − ψm
(b) ψw = ψ p + ψs + ψm (d) None of these
54. Hydrogen bonds occur between which of the following constituents of DNA? (a) Sugar and base (b) Phosphate and base (c) Complementary bases (d) Phosphate and sugar
55. Which one of the following neurotransmitters is secreted by the preganglionic neurons of sympathetic nervous system? (a) Epinephrine (c) Dopamine
(b) Acetylcholine (d) Norepinephrine
56. Which of the following statements about evolution is incorrect? (a) Evolution is the product of natural selection (b) Evolution is goaloriented (c) Prokaryotes evolve faster than eukaryotes
57. Which one of the following compounds is generally translocated in the phloem? (a) Sucrose (c) Dmannose
(b) Dglucose (d) Dfructose
58. Which organelles would be more prominent in a secretory cell than in a nonsecretory cell? (a) Golgi bodies (c) Mitochondria
(b) Lysosomes (d) Pinocytic vesicles
59. In the conversion of RuBP to GP (PGA) (a) a molecule of carbon dioxide is accepted (b) a stable sixcarbon molecule is produced (c) ATP is generated (d) hydrogen is combined with oxygen to form water
60. Adventitious roots develop in (a) creepers (c) twinners
(b) trailers (d) All of these
PARTII (2 Marks Questions) PHYSICS
MATHEMATICS
66. A thin rod of length f / 3 is placed along the principal
61. Let P (a , b) be a variable point satisfying 4 ≤ a + b ≤ 9 and b − 4ab + a ≤ 0. Let R be the complete equation represented in XYplane in which P can lie, the area of region R is 2
(a)
2π 3
2
2
(b) π
2
(c)
4π 3
(d)
5π 3
62. Let S1 (n ) be the sum of first n terms of arithmetic progression 8, 12, 16, ... and let S2 (n ) be the sum of the first n terms of arithmetic progression 17, 19, 21, ... if for some value of n , S1 (n ) = S2 (n ), then this common sum is (a) 216 (c) 200
axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. Linear magnification obtained is 3 2 5 (d) 2
(a) 1 (c)
(b)
1 2
67. Cube of side a is located in three dimensional cartesian space as shown in the figure given below. z
(b) 260 (d) None of these
B
C
63. On a card, the following three statements are found A
1. on this card exactly one statement is false
P
2. on this card exactly two statement are false 3. on this card exactly three statement are false (b) 1
(c) 2
G
(d) 3
64. ∆ABC is right angled at A. The circle with centre A and radius AB cuts BC and AC internally at D and E respectively. If BD = 20 and DC = 16, then the length AC equals (a) 6 21
(b) 6 26
(c) 30
(d) 32
65. The coefficient of x30 in the expansion of (1 + 2x + 3x2 + K + 20x19 + 21x20 )2 is (a) 2706
(b) 2450
(c) 1481
(d) 256
F
y
Q
The number of false statement on the card is exactly (a) 0
D O
E
x
Unit vector in the direction PQ (from centre of face ABOG to centre of face OGEF) is (a) −
2$j −
$ 2k
$ (b) − 2$j + 2 k 1 $ 1 $ (c) j− k 2 2 1 1 $ (d)− $j + k 2 2
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68. In given circuit, cells have zero internal resistances.
Power dissipated in resistors R1 = 20 Ω and R2 = 20 Ω is
R1
20 Ω
R2
– +
20 Ω + – 10 V
10 V
(a) P1 = 5 W , P2 = 0 W (c) P1 = 5 W , P2 = 5 W
(b) P1 = 10 W , P2 = 5 W (d) P1 = 5 W , P2 = 10 W
69. A piece of wood of mass 0.03 kg is dropped from the top of a 100 m high building. At same instant, a bullet of mass 0.02 kg is fired from the ground, with a velocity of 100 ms −1along same vertical line. Bullet gets embeded in block of wood. Height to which combined system rises above the top of the building is (Take, g = 10 ms−2) (a) 10 m
(b) 40 m
(c) 30 m
(d) 20 m
70. If force F, length L and time T are chosen as fundamental quantities, then mass is −1
−2
−1
(a) [FL T ] (c) [FL T −2]
2
(b) [FL T ] (d) [FL0 T −2]
CHEMISTRY 71. 10 dm3 of an ideal monoatomic gas at 27°C and
101 . × 105 N  m−2 pressure heated at constant pressure to 127°C. Thus, entropy change in JK −1 is (a) 2.422 (c) − 2 .422
(b) 5.98 (d) − 5.981
of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ? (Molar mass of Ag = 1078 . , N = 14, O = 16, Na = 23 and Cl = 35.5) (b) 3.5 g
(c) 7 g
(d) 14 g
73. Which of the following diatomic molecules would be stabilised by removal of an electron? (a) C2 (c) N2
(b) CN (d) O2
74. When calcium carbide is hydrolysed, compound X is formed as a major product, which then reacts with dilute sulphuric acid in the presence of mercuric sulphate, a compound Y is formed. Compounds X and Y are (a) C2H2 and CH3 CHO (b) CH4 and HCOOH (c) C2H4 and CH3 COOH (d) C2H2 and CH3 COOH
A
B
Only one structure is possible for B. Identify A, B and C in the reaction (a) CH3 C(CH3 )2 CH3 , CH3 C(CH3 )2 CH2Cl, CH3 C(CH3 )2 CH2CH2C(CH3 )2 CH3 (b) CH3 CHCH2CH3 , CH3 C(Cl)(CH3 )CH2CH3 CH3 , CH2C(CH3 )2 C(CH3 )2 CH2CH3 (c) Both (a) and (b) (d) None of the above
BIOLOGY 76. Which one of the following statements regarding plant growth hormones is correct? (a) Gibberellins do not play any role in flowering (b) Auxin and cytokinin inhibit cell division (c) ABA inhibits root growth and promotes shoot growth at low water potential (d) ABA promotes leaf senescence independent of ethylene
77. A woman with one gene for haemophilia and a gene for colourblindness on one of the Xchromosomes marries a normal man. How will the progeny be? (a) All sons and daughters haemophilic and colourblind (b) 50% haemophilic colourblind sons and 50% colourblind carrier daughters (c) 50% haemophilic daughters and 50% colourblind daughters (d) Haemophilic and colourblind daughters
78. Which one of the following relationships is true in water at 25°C?
72. What is the mass of precipitate formed when 50 mL
(a) 28 g
Na/ ether Light 75. C5 H12 + Cl2 → C5 H11Cl → C
(a) [OH− ] = [H2O− ] −14
(c) Kw > 1 × 10
(b) [H+ ] = [H2O]
(d) [H+ ] = [OH− ]
79. Identify the correct match between the animal (flatworm, earthworm, roundworm) and its body cavity type (acoelomate, coelomate, pseudocoelomate). (a) Roundworm–Pseudocoelomate; Earthworm–Acoelomate; Flatworm–Coelomate (b) Roundworm–Acoelomate; Earthworm–Coelomate; Flatworm–Acoelomate (c) Roundworm–Pseudocoelomate; Earthworm–Coelomate; Flatworm–Acoelomate (d) Roundworm–Coelomate; Earthworm–Pseudocoelomate; Flatworm–Acoelomate
80. A sequence of amino acids may end in either an amino group (—NH2 ) or a carboxyl group (—COOH). What is the theoretical number of chemically different dipeptides that may be assembled from 20 different amino acids? (a) 40
(b) 80
(c) 160
(d) 400
173
KVPY Practice Set 1 Stream : SA
Answers PARTI 1 11 21 31 41 51
(d) (c) (a) (d) (a) (c)
(b)
2 12 22 32 42 52
(d)
62 72
(c)
(c) (c) (a) (b)
(b)
3 13 23 33 43 53
(b)
63 73
(d)
(c) (a) (c) (a)
(b)
4 14 24 34 44 54
(c)
64 74
(a)
(d) (b) (a) (b)
(a)
5 15 25 35 45 55
(b)
65 75
(a)
(c) (a) (c) (b)
(a)
6 16 26 36 46 56
(b)
66 76
(d)
(b) (c) (c) (c)
(b)
7 17 27 37 47 57
(a)
67 77
(b)
(a) (b) (a) (b)
(b)
8 18 28 38 48 58
(a)
68 78
(d)
(a) (d) (d) (b)
(a)
9 19 29 39 49 59
(a)
69 79
(c)
(d) (d) (a) (b)
(a)
10 20 30 40 50 60
(a)
70 80
(d)
(c) (c) (c) (c)
PARTII 61 71
(d) (a)
(b)
(c)
(b)
(a)
(b)
(b)
(a)
(b)
(b)
Solutions 1. (d) We have, x = 3 + 1
πx3 πx3 : 4 12 π π = 1: : 4 12 22 22 : = 1: 28 84 = 42 : 33 : 11
∴V1 : V 2 : V3 = x3 :
Both sides squarring, we get x2 = 3 + 2 3 + 1 = 4 + 2 3 ⇒ 4 4 4 = = ⇒ 2 4 + 2 3 2 ( 2 + 3) x = 2(2 −
3) = 4 − 2 3 2
4 ⇒ x4 + 4 = x2 + 2 − 8 x x 16
4. (b) Let the observation are
= (4 + 2 3 + 4 − 2 3 ) 2 − 8 = (8)2 − 8 = 64 − 8 = 56
2. (b) Time taken by Ajay and Vijay to meet first time anywhere on the path Distance 12 . = = = 0.6 h Relative speed 8.6 Time taken by Ajay and Sanjay to meet anywhere Distance 1.2 = = = 0.8 h Relative speed 9 + 6 The number of times Ajay and Sanjay meets anywhere on the path by the time Ajay and Vijay meets each other for the 36 1 first time = = 7 , i.e. 7 times. 4.8 2
3. (b) Let the side of cube = x x ∴ Radius of cylinder = Radius of cone = 2 Height of cylinder = Height of cone = x Volume of cube (V1 ) = x3 Volume of cylinder 2
x πx (V 2 ) = π ⋅ x = 2 4
3
(V3 ) =
n
∴
x=
∑ xi
i =1
n When 10 is subtracted from each observation, then mean = x − 10 ∴ x − 10 = 60% of x 60 ⇒ x − 10 = x 100 3 ⇒ x − x = 10 ⇒ x = 25 5 When 5 is added to each observation, then new mean is 25 + 5 = 30
5. (a) Given, ∠COA = 60° AO = OB = OC = r In ∆AOC, AO = CO = r and ∠ COA = 60° D
B r l
60°
O
60°
r d
C
A
Volume of cone 2
x1 , x2 , x3 , K , xn
πx 1 x π ⋅x = 3 2 12
3
∴∆AOC is an equilateral triangle. ∴ AC = d = r
DB = DC = l [Q two tangents are equal from external points] 1 ⇒ ∠BOD = ∠COD = ∠BOC 2 ⇒ ∠BOC = 180° − ∠COA = 180° − 60° = 120° 1 1 ⇒ COD = ∠BOC = × 120° = 60° 2 2 CD In ∆OCD, tan ∠COA = OC l tan 60° = ⇒ d l=d 3 ⇒
6. (a) Let y2 = x4 + x3 + x2 + x + 1 2
x x2 Consider x2 + = x4 + x3 + 2 4 3 2 4 3 2 = x + x + x + x + 1 − x + x + 1 4 1 = y2 − (3x2 + 4x + 4) 4 As discriminant of 3x2 + 4x + 4 is negative. ∴ 3x3 + 4x + 4 > 0 2
x Thus, x2 + < y2 2 x2 + x  y > 2 x 1 But x2 + = x + x is nonnegative, 2 2
⇒
∀x ∈ I x2 + x = x2 + x <  y 2 2
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KVPY Practice Set 1 Stream : SA
If x is even, then y≥ x2 + ⇒
x +1 2
y2 ≥ x4 + x3 + x2 + x + 1 +
5 2 x 4 which is not possible. If x ≠ 0, then x = 0 is the only solution when x is even. x 1 If x is odd, then x2 + + is an integer. 2 x x 1 2 So, y≥ x + + 2 2 ⇒
2f (xy) = {f (x)} + {f ( y)} Putting y = 1 ∴ 2f (x) = f (x) + {f (1)}x f (x) = 2x ∴ f (5) = 25 = 32 and f (3) = 23 = 8 f (5) − f (3) = 32 − 8 = 24 y
5 2 x 4
y2 ≥ y2 +
In this case,
x2 x 3 y2 ≥ x4 + x3 + x2 + x + 1 + − − 4 2 4 2 x x 3 i.e. y2 ≥ y2 + − − 4 2 4 1 = y2 + (x2 − 2x − 3) 4 1 and hence (x2 − 2x − 3) ≤ 0 4 x2 − 2x − 3 ≤ 0 x ∈ [−1, 3] ∴ There are exactly 3 integer 0, −1and 3 for which the expression is perfect square and sum = 0 − 1 + 3 = 2 100a + 10b + c Given, 100a + 10b + c = 11(a 2 + b2 + c2 ) ...(i) (99a + 11b) + (a − b + c) = 11(a 2 + b2 + c2 ) 99a + 11b is divisible by 11. ∴a + b + c is must divisible by 11. Hence, so, a − b + c −8 ≤ a − b + c ≤ 18 We conclude a − b + c is either 0 or 11. Now, putting b = a + c in Eq. (i), we get 100a + 10(a + c) + c = 11 [a 2 + (a + c)2 + c2 ] ⇒ 2a 2 + (2c − 10) a + 2c2 − c = 0 ∴The first two terms of this expression are even third term should be even as well ⇒ c is even D = (2c − 10)2 − 4 × 2(2c2 − c) = 4(−3c2 − 8c + 25) is a square of c = 0 When c = 0 ∴ 2a 2 − 10a = 0 ⇒ a = 5, a ≠ 0 b = a + c⇒b = 5 + 0 = 5 ∴Number are 550 Now, when b = a + c − 11 ∴2a 2 + (2c − 32a ) + 2c2 − 23c + 131 = 0 D = 4(−3c2 + 14c − 16) is square c = 3 Q 2a 2 − 26a + 80 = 0, a = 5, a ≠ 8 ∴ b= a+ c= 5+ 3= 8 ∴ Number are 803 ∴ Sum = 550 + 803 = 1353
x
9. (a) M1 = 12 boys, D1 = 15 days, R1 = Rate of working M2 = 9 boys, D2 = 15 days, R2 = Rate of working ∴ M1 D1 R1 = M2D2R2 ⇒ 12 × 15 × R1 = 9 × 15 × R2 R1 9 3 = = ⇒ R2 12 4 R2 4 = ⇒ R1 3 (R − R1 ) Percentage increase = 2 × 100 R1 R = 2 − 1 × 100 R1 4 = − 1 × 100 3 1 = × 100 3 1 = 33 % 3
7. (b) Let threedigits number are
5π 7 BC = AC ∴ ∠BAC = ∠BCA 1 5π π = π − = 2 7 7 4π 3π + = π ∴ ∠GAC + ∠GAD′ = 7 7 Hence, CAD′ are collinear. π ∠GCA = ∠GD ′ A = 7 ⇒ ∠CAB = ∠ACB ∴ ∆GCD′ ~ ∆BAC GC CD ′ GD ′ = = BA AC BC AC CD ′ CA + AD ′ = = ⇒ BA GC AD [QGC = GD = AD ] AC AC + AD = ⇒ AB AD AC + AD 1 1 1 ⇒ = + = AB AC ⋅ AD AC AD 1 1 1 [Q AB = 1] + = =1 ⇒ AC AD AB ∠ABC =
8. (b) We have,
12. (c) We have, radius of circle are 3, 2 and 1.5, respectively. A O 1 R 2 P
10. (a) We know, x x K − 1 = K + 1 when K ≥ 2 Then, x =
K2 − 4 when K is even and 4
K2 − 5 is when K is odd. 4 x x ∴ 100 − 1 = 100 +
1002 − 4 4 = (25)2 − 1 = 2500 − 1 = 2499
∴Positive integers of x =
⇒ In ∆SMT ,
F
S 4 Q
OR = OP − RP OR = 3 − 2 = 1[Q RP = SQ = 2] OS = 3 + 2 = 5 RS 2 = PQ 2 = OS 2 − OR 2 PQ =
ST = 2 + 15 . = 3.5
E′
= 12.25 − 6.25 = 6 MT =
D
D′ A B
B′
C′
24 = 2 6
MT 2 = ST 2 − SM 2 = (3.5)2 − (2.5)2
G
C
T
SM = QM − SQ = 4.5 − 2 = 2.5 [QQM = AP − BT ]
11. (c) Reflect the heptagon with AG as an axis to obtain another heptagon AB ′C ′ D ′ E ′ F ′G ′ F
3
M 2
3
= (5)2 − (1)2 = 25 − 1 = 24
K = 100 which is even.
E
B
In ∆ORS, ⇒ ⇒ ∴
1
N
∴
6
AB = AN + NB = PQ + MT =2 6+
6=3 6
175
KVPY Practice Set 1 Stream : SA 13. (c) Given, 4 × 4 × 4 cubes is made 64 faces 1 × 1 × 1 cubes. Total cubes = 64, White = 20, Red = 44 To find minimum number of visible white box counting total visible faces of unit cube. Total number of faces of small cube on bigger cube except boundary cubes = 4 × 6 = 24 Counting boundary cubes = 16 + 8 + 8 = 32 ∴Total visible faces = 56 But we have 44 red cube. ∴Minimum number of white faces cubes which are visible = 56 − 44 = 12
14. (d) Let x minute will be taken. In 1 one minute A can fill the part of 60 tanker and in one minute B can fill the 1 part of tanker. 40 Both can fill in t minute t t + =1 60 40 ⇒ t = 24 min 1 Both can fill in one minute part of 24 tanker x x 1 1 1 = + ⋅ 2 40 2 24 x 1 1 + =1 2 40 24 120 × 2 x 3 + 5 ⇒ = 1⇒ x = 2 120 8
⇒
⇒
x = 30 min
1 and cos (α − β ) = 1 2 π α+β= 3
⇒ cos (α + β ) = ∴
16. (b) dU =
1 mgx ⋅ A ⋅ dx 2 9A
1 m2 g 2A 2 l2A 2Y m2 g 2l . = 6AY
l
∫0
x2 dx
17. (a) Ice melts at lower temperature due to increase in pressure. As wire passes, the water formed is again freezes and hence wire passes without cutting ice.
18. (a) Conservation of energy gives,
⇒
∴
∆TN 2 ∆THe
1 mu 2 = ∆U 2 1 mu 2 = nCV ∆T 2 mu 2 nMu 2 Mu 2 = ∆T = = 2CV 2nCV 2nCV 3 (MN 2 . CV He ) 14 2 R = = × 5 (MHe ⋅ CV N ) 4 R 2 2 21 2 = ≈ 10 1
19. (d) We plot process along with isotherms. p T2
T1
T3 T4
15. (c) We have, 3 2 1 + cos 2α 1 + cos 2β 3 + = ⇒ 2 2 2 ⇒ cos 2α + cos 2β = 1 ⇒ 2 cos (α + β ) cos (α − β ) = 1 1 ...(i) cos (α + β ) cos (α − β ) = ⇒ 2 1 and sin α sin β = 4 1 ⇒ 2 sin α sin β = 2 1 ...(ii) cos (α − β ) − cos (α + β ) = ⇒ 2 From Eqs. (i) and (ii), we get cos (α + β ) cos (α − β ) − cos (α − β ) + cos (α + β ) = 0 ⇒
⇒
2
U =
⇒
21. (a) By momentum conservation,
cos2 α + cos2 β =
V
Clearly, temperature initially increases then decreases.
20. (c) Linear momentum is conserved in the horizontal direction. ⇒ mv0 sin 37° + 0 = mV − mv sin 37° Along common normal, e(v0 − 0) = V sin 37° + v 27 11 V = v0 and v = v0 ⇒ 34 34 So, impulsive tension = mv0 cos 37° + mv cos 37° 4 11 4 × = mv0 + mv0 34 5 5 18 = mv0 17
4v = (A − 4)v′ 4v v′ = A−4
22. (c) Heat required to melt ice = Q1 = mL = 30 × 80 = 240 cal Heat taken by water formed to reach at 100°C, Q2 = ms∆T = 3000 cal Heat given by steam on condensation = Q3 = mL = 25 × 540 = 13500 cal As heat taken by ice is less than heat given by steam on condensation. So, resulting mixture is at 100°C. Steam condensed Maximum heat absorbed by ice = Latent heat of vapourization 5400 = = 10 g 50 So, resulting mixture contains (30 + 10 = 40 g) of water and (25 − 10 = 15 g) of steam at 100°C. 23. (a) For a freely falling body, distances travelled in each successive second increases in the ratio of successive odd integers. i.e. s1 :s2: s3 : s4 : s5 : s6 : s7 : : 1 : 3 : 5 : 7 : 9 : 11 : 13 s5 9 13 s5 ⇒ = ⇒ s7 = s7 13 9 24. (b) For successful rotation, vbottom = 5 gl vtop = gl These values are minimum possible values. So, ratio of kinetic energies, 1 2 mvbottom K bottom 2 = = 5 :1 1 2 K top mvtop 2
25. (a) Wooden block sinks, when Density of wood = Density of benzene ρb ρw = ⇒ 1 + rw ∆T 1 + rb ∆T ρb − ρw = 21.7°C ∆T = ⇒ ρw rw − ρb rb So, at 22°C wood block sinks in benzene.
26. (c) m = k ρa gb vc ⇒ ⇒ ∴
[M] = [Ma L−3 a + b + c T −2b − c ] a = 1, b = − 3, c = 6 m ∝ v6
27. (b) After a long time, terminal velocity is attained to kinetic energy is constant and only potential energy is dissipated at a constant rate.
176
KVPY Practice Set 1 Stream : SA
28. (d) 2wx = w − x ⇒ x = l 2
l 6 D
C B A x 2w
w
l — –x 2
29. (d) As total energy of system (U + K ) is negative for all values of r, system is a bound system.
30. (c) From ray diagram, A
S
d
B
Distance, AB = d + d + d = 3d ∴ Distance required = 3 m.
31. (d) Lattice energy is defined as the amount of energy required to completely seperate one mole of a solid ionic compound into gaseous constituent ions. It is directly proportional to the charge of the ions. Thus, greater the ionic charge, larger is the lattice energy. Hence, correct order is Al 2O3 > CaO > MgBr2 > NaCl
35. (c) If the specie is a reducing agent,
39. (a) 1 mole of Al requires
it means it can be oxidised easily thus it should have an oxidation number less than the maximum values of oxidation number. Oxidation number of S in the given species are (i) SO2 x + 2(−2) = 0 x=+ 4 (ii) SO32− x + 3(−2) = − 2 x− 6= − 2 x=+ 4 (iii) H 2 SO4 2(1) + x + 4(−2) = 0 2+ x− 8= 0 x=+ 6 (iv) S2− x=−2 As the maximum value of oxidation number of S is − 2. Thus, H 2SO4 can not act as a reducing agent. 36. (d) The wavelength of light is related hc to its energy by the equation E = λ Given, λ = 45 nm = 45 × 10−9 m [Q 1 nm = 10−9 m] 6.63 × 10−34 Js × 3 × 108 ms−1 Hence, E = 45 × 10−9 m
= 3 × 96500 C 10−3 moles of Al requires = 3 × 96500 × 10−3 C = 3 × 96.5 C [1C = As] = 3 × 9.65 As 3 × 96.5 Time (s) = As = 30 s 9.65 A
= 4.42 × 10−18 J
37. (a) N2 + 3H2 → 2NH3
40. (c) The IUPAC name of the following compound is 5 6
3
2 1
4
3, 4, 4trimethylhex1yne .
41. (a) Aldehydes or ketones having atleast one α −H atom undergo aldol condensation. The structures of given compounds are as follows: (a)
HCH
(c) H3C—CH—CH
O
CH3 O 2methyl pentanal (1αH atom)
Methanol (No αH atom)
O (b)
Ph (d) CH2—C—CH3 O
Cyclohexanone (4αH atoms)
1phenyl propanone (5αH atoms)
Thus, among the given compounds, methanal has no αH atom. Hence, it will not give aldol condensation.
= 3.72 × 10−20 cm3
6 g of hydrogen produces 34 g of NH3 ∴450 g of hydrogen produces 34 = × 450 = 2550 g of NH3 6 Actual ammonia produced in the solution = 1575 g 1575 × 100 ∴ % yield = 2550 = 61.76 % ≈ 61.8%
42. (b) Compound II is most basic among the given compounds. This is because the lone pair present on nitrogen in pyridine does not take part in delocalisation and hence they are available for donation. Whereas, in compound I and III the lone pair on N atom takes part in resonance and will not be available for donation, so their basicity will be less.
33. (c) Intensive properties are those
38. (d) The possible isomer of C2BrClFI
43. (a) In Clemmensen reduction,
properties which do not depend upon the quantity or size of matter. Among the given quantities, boiling point (I), pH (III) and emf (IV) are intensive properties whereas entropy (II) is an extensive property.
are as follows Br (i) C==C I
34. (a) Number of radial nodes
(iii)
32. (a) One mole of ideal gas at STP (22.4 L) contains 6.022 × 1023 atoms i.e., 6.022 × 1023 atoms are present in 22400 mL. ∴ Average volume per molecule 22400 = cm3 6.022 × 1023
n = 3, l = 0 ∴ Number of radial nodes = 3 − 0 − 1= 2 For 2porbital,n = 2, l = 1 Number of radial nodes = 2 − 1 − 1= 0
(iv)
F C==C
I
I C==C
I
Br
I
Br (v)
C==C Cl
F
C==C F
F
Br (ii)
Cl
Br
= n − l−1 For 3 sorbital
Cl
Cl I
Br (vi)
C==C
Cl F F Therefore, the above compound has 6 isomers.
Cl
ketones are reduced to alkanes with the help of ZnHg in the presence of HCl. For example,
H3C H3C
C==O
ZnHg HCl
H3C H3C
CH2
44. (b) Higher the reduction potential of a metal, lesser will its reducing power. As the reduction potentials of a metal is decreasing in the order Y > X > Z, thus the reducing power will decrease in the order Z > X > Y.
177
KVPY Practice Set 1 Stream : SA 45. (b) When white phosphorus is heated
53. (b) Water potential (ψw ) is actually
with caustic soda , then sodium hypophosphite with phosphine is formed.
determined by taking into account factors like osmotic (or solute) potential (ψs ) , pressure potential (ψ p ) and matrix or capillary potential (ψm ). The formula for calculating water potential is ψw = ψs + ψ p + ψm
4P
White phosphorus
+ 3NaOH + 3H2O → Caustic soda
NaH2PO2 Sodium hypophosphite
+ PH3 Phosphine
46. (c) The filtration of blood at the glomerulus is a pressure dependent physical process known as renal ultrafiltration. The force of hydrostatic pressure in the glomerulus (the force of pressure exerted from the pressure of the blood vessel itself) is the driving force that pushes filtrate out of the capillaries and into the slits in the nephron.
47. (b) Grave’s disease is an immune system disorder that results in the overproduction of thyroid hormones (hyperthyroidism). Its symptoms include anxiety, irritability, tremor, heat sensitivity, enlargement of thyroid gland, change in menstrual cycle, etc.
48. (b) The ‘imperfect fungi’ belongs to class Deuteromycetes. They are called as imperfect because sexual reproduction is absent in these forms. They reproduce only by asexual spores called conidia.
49. (b) Presence of amniotic egg is not a characteristic of phylum Chordata. The four features shared by all chordates are presence of a single notochord, a dorsal hollow nerve cord, pharyngeal slits and a postanal tail. Amniotic eggs are present in reptiles, birds and mammals only.
50. (c) TCA cycle or Citric acid cycle is a series of reactions that produces one GTP or ATP as well as three NADH molecules and one FADH2 molecule in each turn, which will be used in further steps of cellular respiration to produce ATP for the cell. 51. (c) During anaphaseI, no crossing over leads to segregation of alleles. AnaphaseI begins when the two chromosomes of each bivalent (tetrad) separate and start moving toward opposite poles of the cell as a result of the action of the spindle, but their centromeres are still attached.
52. (d) After vigorous exercise, excess oxygen is required by the body to make ATP for gluconeogenesis, to metabolise lactic acid, to replenish phosphocreatine and glycogen and to pay back any oxygen that has been borrowed from haemoglobin.
54. (c) Hydrogen bonding in DNA occurs between complementary bases in order to keep the two strands of DNA helix together. These bonds occur as 2 hydrogen bonds between adenine and thymine and 3 hydrogen bonds between cytosine and guanine. 55. (b) Both sympathetic and parasympathetic preganglionic neurons are cholinergic meaning they release Acetylcholine (Ach) at the synapse in the ganglion. Ach basic functions involve the control of skeletal muscles via activation of the motor neurons as well as stimulating the muscles of the body.
56. (b) Evolution is not goal oriented. Evolution simply depends on the environment, which the organisms live and try to survive. The environment is fit for strongest individual who can survive and reproduce. Evolution uses the theory of natural selection where there is variation. We have variations of traits, heredity and different reproductive strategies as a result of natural selection.
57. (a) The sucrose is actively transported against its concentration gradient into the phloem cells using the electrochemical potential of the proton gradient. This is coupled to the uptake of sucrose with a carrier protein called the sucroseH+ symporter.
58. (a) A secretory cell would need secretory enzymes and glycoproteins required in secretions, which are produced in Golgi bodies.
Y
60° X O 15° (2,0) (3,0)
X′
Y′
b Let = tan θ a 2 b − 4b + 1 ≤ 0 a a ⇒ ⇒ ⇒
tan 2 θ − 4 tan θ + 1 ≤ 0 tan θ ∈ [2 − 3 , 2 + 3 ] θ ∈ [15° , 75° ] π Area of region = (32 − 22 ) 3 π 5π = (9 − 4) = 3 3
62. (b) Given, S1 (n ) = 8 + 12 + 16 + K + n terms n n S1 (n ) = [16 + (n − 1) 4] = (12 + 4n ) 2 2 and S2 (n ) = 17 + 19 + 21 + K+ n terms n n S2 (n ) = [34 + (n − 1) 2] = (32 + 2n ) 2 2 S1 (n ) = S2 (n ) n n Q (12 + 4n ) = (32 + 2n ) ⇒ n = 10 2 2 Q S1 (10) = 5(16 + 36) = 260 = S2 (10) Q Common sum = 260
63. (c) If any statement is true, then remaining 2 are false. 64. (b) Given, ∆ABC is right angled at A. A is centre of circle and AB is radius of circle. BD = 20 CD = 16 C 16
59. (a) Riboluse 1,5Biphosphate (RuBP)
E
is the first acceptor of CO2 in the formation of two molecules of 3Phosphoglyceraldehyde (PGA) during the Calvin cycle of photosynthesis.
r
60. (a) Horizontal stem of creepers often develop adventitious roots from the nodes. Adventitious roots are the roots which arise from an organ other than a root. They generally develop from stem nodes, internodes, leaves, etc.
61. (d) Given, 4 ≤ a 2 + b2 ≤ 9 and
b2 − 4ab + a 2 ≤ 0
A
D 20 r
B
r F
In ∆ABC, AC 2 + r 2 = BC 2 AC 2 + r 2 = (36)2 CB and CF are secant of circle. Q CE − CF = CD × CB ⇒ (AC − r ) (AC + r ) = 16 × 36 AC 2 − r 2 = 16 × 36
...(i)
...(ii)
178
KVPY Practice Set 1 Stream : SA
From Eqs. (i) and (ii), we get 2AC 2 = 36 (36 + 16) = 36 × 52 AC = 6 26 ⇒
Time at which bullet and block collide is d 100 t= = = 1s v 100 − 0
65. (a) We have,
Speed of wood just before collision = gt = 100 ms−1 Speed of bullet at t = 1s is = v − gt = 100 − 10 × 1 = 90 ms−1 . Let v′ = velocity of bullet + block system after collision. Then, by momentum conservation, we have − (0.03) × 10 + (0.02) × 90 = (0.05) v′ ⇒ v′ = 30 ms−1 Now, maximum height reached v2 = − distance through 2g which block fells in 1 s 30 × 30 1 2 = − × 10 × 1 = 45 − 5 2 × 10 2
(1 + 2x + 3x + K + 21x ) = (1 + 2x + 3x2 + K + 21x20 ) (21x20 + 20x19 + K + 3x2 + 2x + 1) Coefficient of x30 is 11 × 21 + 12 × 20 + ... + 21 × 11 = 2(11 × 21 + 12 × 20 + 13 × 19 + 14 × 18 + 15 × 17) + 16 × 16 = 2(231 + 240 + 247 + 252 + 255) + 256 = 2(1225) + 256 = 2450 + 256 = 2706 2
20 2
66. (b) Given situation is
B′
A′ A
B
C
F
By mirror formula, 1 1 1 5f − = − ⇒ vB = − 5 vB f 2 f 3 5 f Hence, image length is l′ = f − 2 = 2 2 l′ f / 2 3 So, magnification is m = = = l f /3 2
67. (b) Coordinates of points P and Q are a a a a P : , 0, and Q : , , 0 2 2 2 2 So, unit vector along PQ is a a$ − $j + k PQ ^ 2 2 PQ = = PQ a2 a2 + 4 4 $ $ = 2 (− j + k )
68. (a) By KVL, I1 =
10 20
100 × 20 = 5 W 400
⇒
P1 = I12R1 =
and
I 2 = 0 ⇒ P2 = 0
69. (b) u=0
100 m
v=100 ms–1
= 40 m
70. (b) M = kF a LbT c where, k is a constant. [M] = [MLT −2 ]a Lb T c Equating dimensions, we have …(i) a=1 …(ii) b+ a = 2 …(iii) −2a + c = 0 Putting value of a from Eq. (i) in Eq. (ii), we get b+ a = 0 b+ 1= 0 …(iv) b = −1 Again putting value of a from Eq. (i) in Eq. (iii), we get −2a + c = 0 −2 × 1 + c = 0 −2 + c = 0 c=2 ⇒ a = 1, b = −1 and c = 2 So, [M] = [FL−1 T 2 ]
72. (c) 16.9% solution of AgNO3 means 16.9 g AgNO3 is present in 100 mL solution. ∴ 8.45 g AgNO3 will be present in 50 mL solution. Similarly, 5.8 g NaCl is present in 100 mL solution. ∴2.9 g NaCl is present in 50 mL solution. Initial moles
After reaction 0
= 0.405 mol
[1N  m =1 J] 5 For monoatomic gases, C p = R = 2 .5R 2 T1 = 300 K and T2 = 400 K R = 8. 314 J mol −1 K −1 T ∆ S = 2. 303 nC p log 2 T1 400 = 2.303 × 0.405 × 2. 5 × 8.314 log 300 ∆S = 2. 422 JK −1
0
0.049 0.049
Mass of compound = moles × molar mass ∴ Mass of AgCl precipitated = 0.049 × 143.5 = 7.03 g
73. (d) The species formed after removal of an electron from the given diatomic molecules are as follows (a) C2 → C2+ + e− (b) CN → CN+ + e− (c) N2 → N2+ + e− (d) O2 → O+2 + e− The stability of diatomic molecule can be determined by calculating its bond order. More is the bond order, more is the stability of a molecule. 1 B.O = (Nb − N a ) 2 ∴ The bond orders of diatomic molecules with their ionic species are given below. (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
71. (a) From ideal gas equation, pV = nRT pV 1. 01 × 105 N  m −2 × 10 × 10−3 m3 = n= RT 8. 314 J mol −1 K −1 × 300 K
AgNO 3 + NaCl → AgCl + NaNO 3 8.45 2.9 0 0 169.8 58.5 = 0.049 = 0.049
C2 C+2 CN CN+ N2 N2+ O2 O 2+
B.O 2.0 1.5 2.5 2.0 3 2.5 2 2.5
As the bond order increases from 2 to 2.5 when an e− is removed from O2 molecule (O+2 ), so it become stabilised.
74. (a) Calcium carbide on hydrolysis gives acetylene as a major product which then reacts with dil H2SO4 in the presence of HgSO4 to give acetaldehyde. CaC2 + 2H2O → C2H2 + Ca(OH)2 Calcium carbide
(X) Acetylene
CH dil H 2 SO4 / HgSO4 → CH
CH3 C H O (Y ) Acetaldehyde
179
KVPY Practice Set 1 Stream : SA 75. (a) As only one structure of B with
76. (d) ABA or Abscisic Acid promotes
molecular formula C5 H11 Cl is possible, thus the structure of B would be CH3 CH3 C CH2 Cl. CH3
leaf senescence independent of ethylene. Other statements can be corrected as Gibberellins play the most important role in flowering. It is seen that treatment of gibberellin on biennials or long day plants, stem elongation occurs before flower primordia are formed. Auxin and cytokinin promote cell division. It is seen that ABA and ethylene can control and induce root and shoot growth under water stress or low water potential.
CH3 Light CH3 C CH3 + Cl 2 ← CH3 ( A)
CH3 CH3 C CH2 Cl CH3 (B ) Wurtz reaction
Na / ether ↓
CH3 CH3 CH3 C CH2 CH2 C CH3 CH3 CH3 (C)
77. (b) The cross for the given question would be XhcX XhcX Colourblind haemophilic carrier girl
XhcY
×
X Y
XX
Colourblind Normal girl haemophilic boy
XY Normal boy
∴ The progeny be 50% haemophilic colourblind sons and 50% colourblind carrier daughters.
78. (d) In pure water, the concentration of hydrogen ions and concentration of hydroxyl ions are equal. Consequently
water is neither acidic nor basic, but neutral. Therefore, water at 25°C will have H+ = OH− .
79. (c) Acoelomate Animals like sponges, coelenterates and flatworms are without a coelom or any other internal cavity except the digestive tract. Pseudocoelomate Roundworms have a body cavity derived directly from the blastocoel of the embryo. It is called pseudocoel because like true coelom, it is not lined by peritoneum but is bounded with ectoderm on the outer side and endoderm on the inner side. Coelomate Animals with a tube within a tube body plan have a fluidfilled body cavity between the body wall and the digestive tract as like in earthworms. It is derived from embryonic mesoderm and is lined by peritoneum. 80. (d) A dipeptide is made up of 2 amino acids which may be same or different. The total possible number of different dipeptides that may be assembled from 20 different amino acids will thus be 20n = 202 = 20 × 20 = 400
180
KVPY Practice Set 2 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
PRACTICE SET 2 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS 1. The remainder when 5 (a) 6 (c) 9
4. Difference between the corresponding roots 99
is divided by 13 (b) 8 (d) 10
2. A polynomial p(x) when divided by x2 − 3x + 2 leaves remainder 2x − 3, then (a) p(x) must have a root between 0 and 3 (b) p (x) cannot have a root between 0 and 3 (c) p (x) must have a real root but may or may not be between 0 and 3 (d) p (x) need not have a real root
3. A shopkeeper increases the price of a commodity by x% some time later, he reduces the new price by y% and notices that the price is now the same as it was 1 1 originally. The value of − is y x 1 100 1 (c) 100
(a) −
(b) 0 (d) None of these
x2 + ax + b = 0 and x2 + bx + a = 0 is same and a ≠ b, then (a) a + b + 4 = 0 (c) a − b − 4 = 0
(b) a + b − 4 = 0 (d) a − b + 4 = 0
5. Let P be a point in the interior of the rectangle ABCD, which of the following sets of numbers can form the areas of the four triangles PAB, PBC, PCD, PDA in same order (a) 10, 9, 12, 5 (c) 10, 9, 8, 6
(b) 21, 15, 6, 12 (d) 12, 8, 7, 5
6. Let x1 , x2 , K , xn be n observation such that n
∑
i =1
xi2 = 400 and
n
∑
i =1
xi = 80. Then, a possible value of
n among the following is (a) 15
(b) 18
(c) 9
(d) 12
7. The set S = {1, 2, 3, K , 12} is to be partitioned into three sets A, B, C of equal size. Thus, A ∪ B ∪ C = S, A ∩ B = B ∩ C = A ∩ C = φ. The number of ways of partition S is (a)
12! 3! (4!)3
(b)
12! 3! (3!)4
(c)
12! (4!)3
(d)
12! (3!)4
181
KVPY Practice Set 2 Stream : SA 8. You have a measuring cup with capacity 25 ml and another with capacity 110 ml, the cups have no markings showing intermediate volumes. Using large container and as much tap water as you wish. What is the smallest amount of water you can measure accurately? (a) 1 ml (c) 10 ml
PHYSICS 16. A box is falling freely inside the box, a particle is projected with some velocity v with respect to the box at angle θ.
(b) 5 ml (d) 25 ml
v
9. Let A, B, C, D be collinear points in that order.
Suppose AB : CD = 3 : 2 and BC : AD = 1 : 5. Then, AC : BD is
(a) 1 : 1 (c) 16 : 1
(b) 11 : 10 (d) 17 : 13
10. Let ABC be triangle with AB = AC = 6. If the circumradius of the triangle is 5, then BC equals 25 3 48 (c) 5
(d) 10
17. Potential energy of a system as a function of a
the same place. The first goes with a speed of 10 km/h. The second goes at a speed of 8 km/h in the 1 first hour and increase the speed by km each 2 succeeding hour. After how many hours will the second car overtake the first, if both go nonstop. (a) 9 h (c) 7 h
(b) 5 h (d) 8 h
12. If x, y are natural numbers such that x2 + 2013 = y2, then the minimum value of xy is (a) 645 (c) 668
(b) 658 (d) 671
13. A cube has each edge 2 cm and a cuboid is 1 cm long, 2 cm wide and 3 cm high. The paint in a certain container is sufficient to paint an area equal to 54 cm 2. Which one of the following is true? (a) Both cube and cuboid are painted (b) Only cube can be painted (c) Only cuboid can be painted (d) Neither cube nor cuboid can be painted
14. How many positive real number x are there such that = (x x )x ?
(a) 1 (c) 4
all 3digit number formed by using all the 3digit number once each is 1554. The value of b is (b) 2
(c) 3
parameter x is U (x) = (x + 1) (x + 2). Then, consider following statements: −3 I. Point x = , corresponds to an equilibrium 2 position. II. Points x = − 1 and x = − 2, corresponds to equilibrium position of system. −3 III. At point x = , system is in stable equilibrium. 2 −3 , system is in unstable IV. At point x = 2 equilibrium.
(a) All statements are correct (b) Statements I and IV are correct (c) Statements I and III are correct (d) Only statement II is correct
18. On a temperature scale X, water boils at − 60° X and freezes at − 180° X. What would be a room temperature of 25° C on Xscale? (a) − 18° X
(b) − 38° X
(c) − 150° X
(d) − 130° X
19. In given nuclear reaction, 9 4 Be
+α→
(a) 42 He
12 6C
+ X, particle X is
(b)
0 −1 e
(c) 11 H
(d) 10 n
20. Two blocks of masses 0.2 kg and 0.5 kg are placed
(b) 2 (d) Infinite
15. Let 0 < a < b < c be three distinct digits. The sum of
(a) 1
(c) (d)
11. Two cars start together in the same direction from
x
(b)
(a)
(b) 9
(a)
(x)x
θ
For an observer sitting in the box, path of particle is
(d) 4
22 m apart on a rough flat horizontal surface (µ = 05 . ). At time t = 0, blocks are pushed towards each other with equal forces of 3 N on each of the block. Blocks collide with each other in time duration
(a) 1 s (c) 3 s
(b) 2 s (d) 2 s
182
KVPY Practice Set 2 Stream : SA
21. A tabletennis ball is floating in air by a jet of water emerging from a nozzle.
(a) 1 (c) 0.25
(b) 0.5 (d) 1.25
26. If a low pressure centre is developed in atmosphere (very common in India in summer), the wind will flow radially towards centre. The whirlpool of wind formed will rotate in India as
h
(a) clockwise only (b) anticlockwise only (c) clockwise or anticlockwise (d) whirlpools are not formed in India
Ground level
27. Which of these paths correctly describes motion of If mass of ball is m and water stream rises to height h above ground, then water flow rate is (a)
m 2
g h
(b)
m 2
(a)
Moon
22. A system under goes three processes listed in table
(d)
∆Q
∆W
∆U
Process 1 → 2
a
100
100
Process 2 → 3
b
− 50
c
Process 3 → 1
100
d
− 200
Then, value of c is (a) 200 kJ (c) 100 kJ
Moon
Moon
(c)
below. All quantities are (in kJ). Process
Sun
(b)
Sun
2h g
(d) m
(c) m 2 gh
h g
moon around observed from a space station?
Sun
Sun
28. Correct variation of velocity of a tabletennis ball dropped from top of a 14story building is (a) v
(b) 50 kJ (d) 0 kJ
(b)
v
23. Force necessary to accelerate a mass of 1 kg at
10 ms−2 vertically upwards is (Take, g = 10 ms−2)
(a) 1 N
(b) 2 N
(c) 10 N
(d) 20 N
t
t
(c) v (d)
24. In Rutherford’s scattering experiment, choose the
v
correct statements are given below. I. Only αparticles are scattered backwards but not protons. II. αparticles cannot be effectively scattered by electrons because αparticles are positively charged. III. Radius nucleus of target is between 96 . × 10−15 m to 48 . × 10−15 m.
t
29. A vessel with water is placed on a weighing pan, it reads 600 g. Now a hollow ball of mass 40 g and volume 50 cm3 is kept immersed in water by tying it to bottom with a thread of negligible mass.
IV. αparticles with energy greater than a certain critical value are not scattered back. (a) Only statement II is correct (b) Only statement III is correct (c) Statements I, III and IV are correct (d) All statements are correct
25. The ratio of the height above the surface of earth to the depth below the surface of earth for gravitational acceleration to be same (assuming small height) is
500 g
Reading of pan is now (a) 690 g (b) 550 g (c) 650 g (d) 610 g
t
183
KVPY Practice Set 2 Stream : SA 30. Plane face of a planoconvex lens is silvered. Given, radius of convex face is 12 cm and refractive index of medium is 3 / 2 .
(b) 3, 5dimethyl1cyclohexene (c) 1, 5dimethyl5cyclohexene (d) 1, 3dimethyl5cyclohexene
36. Which of the following isomerisms is shown by pentan2one and 3methylbutanone? (a) Stereoisomerism (c) Functional isomerism
37. What is the maximum number of orbitals that can be
Power of resulting system is 25 D 3 − 25 m (c) 3
25 D 3 25 m (d) 3
identified with the following quantum numbers?
(b) −
(a)
(b) Position isomerism (d) Chain isomerism
(a) 1 (c) 3
n = 3, l = 1and ml = 0 (b) 2 (d) 4
38. The major product of the following reaction is (i) O
3 CH 3CH== CHCH2CH3 →
( ii ) hydrolysis, Zn
CHEMISTRY 31. By heating 10 g CaCO3 , 5.6 g CaO is formed. The weight of CO 2 obtained in this reaction is closest to (a) 5.6 g (c) 4.4 g
(b) 2.4 g (d) 3.6 g
32. Which of the following plot obeys the Raoult’s law at
(b)
Vapour pressure
(a)
Vapour pressure
all concentration?
Mole fraction of solvent
(a) PH5 and BiCl5 do not exist (b) pπ − dπ bonds are present in SO2 (c) SeF4 and CH4 have same shape (d) I3+ has bent structure
following reaction equilibrium, 2SO2 ( g ) + O2 ( g )
Mole fraction of solvent
(d)
39. Among the following, which is an incorrect statement.
40. Predict the effect of increased pressure on the
Vapour pressure
(c)
Vapour pressure
Mole fraction of solvent
(a) CH3 CHO + CH3 CH2CHO (b) CH3 COOH + CH3 COCH3 (c) CH3 COOH + CH3 CH2COOH (d) CH3 COOH + CO2

2SO3 ( g )
(a) equilibrium shift to the right (b) equilibrium shift to the left (c) no effect on equilibrium (d) reaction stops
41. The solubility product of BaCl2 is 4 × 10−9. Its solubility in mol L−1 is
Mole fraction of solvent
33. The correct order of increasing ionic character is (a) BeCl 2 < MgCl 2 < CaCl 2 < BaCl 2 (b) BeCl 2 < MgCl 2 < BaCl 2 < CaCl 2 (c) BeCl 2 < BaCl 2 < MgCl 2 < CaCl 2 (d) BaCl 2 < CaCl 2 < MgCl 2 < BeCl 2
34. In the reaction, 3Br2 + 6CO32− + 3H2O → 5Br− + BrO3− + 6HCO3− (a) Bromine is oxidised and the carbonate radical is reduced (b) Bromine is reduced and the carbonate radical is oxidised (c) Bromine is neither reduced nor oxidised (d) Bromine is both reduced and oxidised
35. IUPAC name of the following compound
(a) 4 × 10−3 (c) 1 × 10−3
(b) 4 × 10−9 (d) 1 × 10−9
42. Chlorobenzene on treatment with sodium in dry ether gives diphenyl. The name of the reaction is (a) Fittig reaction (b) Wurtzfittig reaction (c) Sandmeyer reaction (d) Gattermann reaction
43. A sample of unknown gas is placed in a 2.5 L bulb at a pressure of 360 torr and at a temperature of 22.5°C and is found to weight 1.6616 g. The molecular weight of the gas is closest to (a) 80 g (c) 34 g
(b) 55 g (d) 55 g
44. Consider the isoelectronic ions K + ,S2− ,Cl− and Ca 2+
is CH3
CH3
(a) 3, 5dimethyl cyclohexene
The radii of these ionic species follow the order (a) Ca 2+ > K+ > Cl − > S2− (b) Cl − > S2− > K+ > Ca 2+ 2− − + 2+ (c) S > Cl > K > Ca (d) K+ > Ca 2+ > S2− > Cl −
184
KVPY Practice Set 2 Stream : SA
45. The reaction of toluene with Cl2 in the presence of FeCl3 gives predominantly (a) benzoyl chloride (b) benzyl chloride (c) o and pchlorotoluene (d) mchlorotoluene
BIOLOGY 46. Most common type of phospholipids in the cell membrane of nerve cell is (a) phosphatidylcholine (c) phosphatidylserine
(b) phosphatidylinositol (d) sphingomyelin
(a) erythrocytes (b) Tcells (c) macrophages (d) polymorphonuclear leukocytes
(b)
x 2
(c) x
(d) 2x
53. Kreb’s cycle was discovered by Krebs in pigeon muscles in 1940. Which step is called gateway step/link reaction/transition reaction in respiration? (a) Glycolysis (b) Formation of acetyl CoA (c) Citric acid formation (d) ETS terminal oxidation
54. Which homeostatic function of the liver is controlled and monitored in the pancreas?
55. During generation of an action potential, depolarisation is due to
48. Horseshoe crabs belong to the group (b) Chelicerata (d) Crustacea
49. The first living being on the earth were anaerobic because (a) there was no oxygen in air (b) oxygen damages proteins (c) oxygen interferes with action of ribozymes (d) they evolved in deep sea
50. The presence of Salmonella in tap water is indicative of contamination with (a) industrial effluents (c) agricultural waste
x 4
(a) Deamination of amino acids (b) Release of glucose (c) Release of iron (d) Release of toxins
47. Graft rejection does not involve
(a) Onychophora (c) Uniramia
(a)
(b) human excreta (d) None of these
51. The secondary order of protein structure is (a) the sequence of amino acids in the polypeptide chain (b) the formation of peptide bonds between amino acids (c) the coiling of the polypeptide chain (d) the folding of the coiled polypeptide chain
52. The amount of DNA in a mammalian cell in early prophaseI is x. What is the amount of DNA in the same cell in anaphaseI of mitosis?
(a) K+ efflux (c) Na + influx
(b) Na + efflux (d) K+ influx
56. If liver from body is removed then which component of blood increases? (a) Ammonia (c) Urea
(b) Protein (d) Uric acid
57. In his classical experiments on pea plants, Mendel did not use (a) seed shape (c) seed colour
(b) flower position (d) pod length
58. In phylum, which group contains the greatest number of species? (a) Class
(b) Family
(c) Genus
(d) Order
59. Cell division is initiated by (a) centrosome (c) centriole
(b) centromere (d) None of these
60. The part of human hindbrain that is responsible for handeye coordination is (a) cerebellum (c) medulla oblongata
(b) pons Varolii (d) thalamus
PARTII (2 Marks Questions) MATHEMATICS
63. If there are three different kinds of mangoes for sale
61. A circle is inscribed in an equilateral triangle with side length of 6 units. Another circle is drawn inside the triangle (but outside the first circle), tangent to the first circle and two of the sides of the triangle. The radius of the smaller circle is (a)
1 3
(b)
2 3
(c)
1 2
(d) 1
62. If x2 y3 = 6. Then, the minimum value of 3x + 4 y for positive values of x and y is (a) 6
(b) 8
(c) 10
(d) 12
in a market. Then, number of ways of purchase of 25 mangoes are (a) 2925 (c) 351
(b) 325 (d) 2600
64. Four natural number m, n if (1 − y)m (1 + y)n = 1 + a1 y + a 2 y2 + K and a1 = a 2 = 10, then (m, n ) is (a) (20, 45) (c) (45, 35)
(b) (35, 20) (d) (35, 45)
185
KVPY Practice Set 2 Stream : SA 65. In a ∆ABC, with ∠A = 90°, the bisector of the angle B and C meet at P. The distance from P to the hypotenuse is 4 2. The distance AP is (a) 8
(b) 4
(c) 8 2
(d) 4 2
PHYSICS 66. A planet contains a single type of gas in its
atmosphere having molecular mass of 138 . × 10−28 kg. Distribution of speeds in atmosphere is given below. Speed (ms−1) 100 200 500 800 1000
When a very distant object is viewed by this lens system, choose the correct option. (a) Final image is formed at midpoint of lens separation (b) Final image is virtual (c) Diverging lens increases the magnification five times (d) Final image is inverted and diminished
70. Current is flowing through a uniform thick rod of crosssectional area A, under an applied potential difference V across its ends.
Percentage of molecules 10 30 20 20 20
vd
A
+ – V
Escape speed for the planet is 900 ms . Assuming stable atmospheric conditions, the possible estimated reduction in temperature of the planet in few years will be (Use, temperature, 2 mvrms and K B = 138 T= . × 10−23 ) 3 KB
Let electrons flow through the thick rod with velocity vd$ . A hole is drilled in the rod and its central portion 1 A of area is removed. 2 Let electrons flow through the hollow rod with velocity vd$ . Then, ratio vd$ / vd$ will be
(a) 100 K
(a) 2
−1
(b) 200 K
(c) 70 K
(d) 20 K
1 (b) 2
67. A right angle ruler used generally in tailoring or drafting hangs from rest from a peg P as shown below. P θ 2L
L
2
2
1 4 1 (c) cos θ = 4
1 4 1 (d) sec θ = 4
(a) tanθ =
(b) sinθ =
68. 1 kg of steam at 100°C and 101 kPa occupies 1.68 m3 space. What per cent of heat of vaporisation of water is used for expansion of water into steam? (a) Nearly 7% (c) Nearly 70%
(b) Nearly 17% (d) Nearly 12%
69. A telephoto lens system consists of a converging lens ( f = + 60 . cm) placed 4 cm in front of a diverging lens (f = − 2.5 cm).
(d) 4
71. A bomb calorimeter has a heat capacity of 783 J°C −1 and contains 254 g of water, which has a specific heat of 4.184 g −1 °C −1. Heat absorbed/evolved by a reaction when the temperature changes from 23.73°C to 2601 . °C is closest to (b) 2.42 kJ absorbed (d) 4.21 kJ absorbed
72. Isostructural species are those species which have the same shape and hybridisation. Among the given species, identify the isostructural pairs. (a) NF3 and BF3 (c) BCl3 and BrCl3
(b) BF4− and NH+4 (d) NH3 and NO3−
73. Which among the following will form geometrical isomers? (a)
(b)
(c)
(d)
74. Consider the following reaction, CH CH 3CH 2OH B ← CH3 C Br → A (Major) (CH 3CH 2OH) (Major) CH (CH 3CH 2O− Na + )
A and B respectively are
4.0 cm
f=6 cm
(c) 1
CHEMISTRY
(a) 1.78 kJ absorbed (c) 1.78 kJ evolved
One arm is L cm long and other arm is 2 L cm long. Value of angle θ is such that
1
f=–2.5 cm
(a) (CH3 )3 COCH2CH3 in both cases (b) (CH3 )2 C == CH2 in both cases (c) (CH3 )3 COCH2CH3 and (CH3 )2 C== CH2 (d) (CH3 )2 C == CH2 and (CH3 )3 COCH2CH3
186
KVPY Practice Set 2 Stream : SA
75. Which of the following species contains equal number of σ and πbonds?
(a) HCO3− (c) (CN)2
77. Which one of the following genotypes cannot occur amongst the offspring from a mating between a person of blood group A and a person of blood group B?
(b) XeO4 (d) CH2 (CN)2
(a) AA
BIOLOGY
(c) AO
(d) BO
from the electron transport chains in photophosphorylation. Why does this kill the plant?
76. A piece of mammalian tissue was homogenised and subjected to differential centrifugation. The diagrams below indicate the relative activity certain biochemical processes in these fractions. Which of the following fractions indicates the maximum hydrolytic enzyme activity? Activity (a)
(b) AB
78. The weed killer DCMU blocks the flow of electrons
(a) Active transport of mineral ions is prevented (b) ATP and reduced NADP are not produced (c) Photoactivation of the chlorophyll cannot occur (d) Photolysis of water does not occur
79. Which one of the following is the correct description of a certain part of a normal human skeleton?
Activity
Nuclei
(b)
Mitochondria
(a) Parietal bone and the temporal bone of the skull are joined by fibrous joint (b) First vertebra is axis which articulates with the occipital condyles (c) The 9th and 10th pairs of ribs are called the floating ribs (d) Glenoid cavity is a depression to which the thigh bone articulates
Nuclei Mitochondria
Lysosomes
Lysosomes
Ribosomes
Ribosomes
Activity
Activity
80. Which of the following statements correctly describes (c)
(d)
Nuclei
a codon? (a) A length of DNA which codes for a particular protein (b) A part of the transfer RNA molecule to which a specific amino acid is attached (c) A part of the transfer RNA molecule which recognises the triplet code on the messenger RNA (d) A part of the messenger RNA molecule that has a sequence of bases coding for an amino acid
Nuclei Mitochondria
Mitochondria Lysosomes
Lysosomes
Ribosomes
Ribosomes
Answers PARTI 1 11 21 31 41 51
(b) (a) (a) (c) (c) (c)
(a)
2 12 22 32 42 52
(c)
62 72
(b)
(b) (c) (c) (a)
(c)
3 13 23 33 43 53
(b)
63 73
(a)
(a) (d) (a) (c)
(a)
4 14 24 34 44 54
(b)
64 74
(c)
(b) (c) (d) (c)
(b)
5 15 25 35 45 55
(c)
65 75
(b)
(b) (b) (a) (c)
(b)
6 16 26 36 46 56
(a)
66 76
(c)
(b) (b) (d) (a)
(c)
7 17 27 37 47 57
(d)
67 77
(a)
(c) (d) (a) (a)
(b)
8 18 28 38 48 58
(a)
68 78
(b)
(c) (d) (a) (b)
(d)
9 19 29 39 49 59
(a)
69 79
(a)
(d) (c) (c) (a)
(c)
10 20 30 40 50 60
(a)
70 80
(c)
(d) (b) (a) (b)
PARTII 61 71
(a) (d)
(c)
(c)
(d)
(a)
(c)
(a)
(a)
(c)
(c)
187
KVPY Practice Set 2 Stream : SA
Solutions 1. (b) 599 = 598 ⋅ 5 = (52 )49 ⋅ 5
⇒α 2 + β 2 − 2αβ = γ 2 + δ 2 − 2γδ
= (25)49 ⋅ 5 = 5(26 − 1)49 = 5(26k − 1)[Q (a − b)n = nk − (b)n] = 5 × 26k − 5 = 5 × 26k − 13 + 8 = 13(10k − 1) + 8 ∴When 599 is divided by 13 the remainder is 8.
⇒ (α + β )2 − 4αβ = (γ + δ)2 − 4γδ
5. (b) P be an interior point of
2. (a) Let
rectangle ABCD.
p (x) = q(x) (x2 − 3x + 2) + (2x − 3) p (x) = q(x) (x − 1) (x − 2) + (2x − 3) p(1) = 0 + (2 − 3) = − 1 p(2) = 0 + (4 − 3) = 1 ∴ p(1) < 0 and p(2) > 0 ∴ p (x) has one root lie between 1 and 2. ∴ p (x) must have a root lie between 0 and 3.
3. (c) Let the original price of commodity =P Price of commodity when price x% increase P (100 + x) = P + x% of P = 100 Price of commodity when priceY % decrease from the increase of x% 100 + x P (100 + x) Y = P × − 100 100 100 P (100 + x) 100 − y = 100 100 Given, the reduces price is equal to original price. 100 + x 100 − y P ∴ =P 100 100 100 + x 100 − y = 1 100 100 x y xy ⇒ 1+ − − =1 100 100 (100)2 x y xy − = ⇒ 100 100 (100)2 xy x− y= ⇒ 100 1 1 1 Divide by xy, we get − = y x 100
⇒
9. (d) Given, ABCD is collinear. A
a 2 − 4b = b2 − 4a
⇒ (a 2 − b2 ) + 4(a − b) = 0 ⇒ (a − b) (a + b + 4) = 0 ⇒
a+ b+ 4= 0
[Q a ≠ b]
D
C
A
B
∴ Area of ∆APB + area of ∆PCD = Area of ∆PBC + Area of ∆PAB ∴ In option (b), 21 + 6 = 15 + 12 ∴ Option (b) is correct.
6. (b) We have, n
xi2 = 400 and
i =1
n
∑
xi = 80
i =1
We know, ⇒
x12 + x22 + x32 + K + xn2 n 2 x + x2 + x3 + K + xn ≥ 1 n Σxi2 (Σxi )2 ≥ n n2 6400 400 (80)2 = 16 ≥ ⇒ n≥ 400 n n2
⇒ ⇒ ∴
2AC − 2BC = 3BD − 3BC ...(i) BC = 3BD − 2AC BC 1 and = AD 5 5BC = AD ⇒ 5BC = AB + BC + CD ⇒ 4BC = AB + CD ⇒ 4BC = AC − BC + BD − BC ...(ii) ⇒ 6BC = AC + BD From Eqs. (i) and (ii), we get 6(3BD − 2AC ) = AC + BD ⇒ 18BD − 12AC = AC + BD ⇒ 13AC = 17BD AC 17 = ⇒ BD 13
10. (c) We have, In ∆ABC, AB = AC = 6 Circumradius (R ) = 5 A
equal size. ∴ Each set has 4 elements. ∴ Total number of ways in partition is
R=
12! 12! 8! 4! = × × 4! 8! 4! 4! 0! 4! (4!)3
x2 + ax + b = 0 ∴ α + β = − a , αβ = b and γ , δ are roots of equation x2 + bx + a = 0 ∴ γ + δ = − b, γδ = a Given, α − β  =  γ − δ  (α − β )2 = (γ − δ)2
container and take y time of water of 25 ml from container. Then, container contains 110x − 25 y = 5(22x − 5 y) ∴Container contains multiple of 5. ∴Smallest amount of water be measure accurately 5 ml.
5 C
a
We know,
C4 × C4 × C4 .
8. (b) Put x time of water of 110 ml to
6
5 B
4
4. (a) Let α , β are roots of equation
5
6
7. (c) The set S is divided into three
8
D
∴AB + BC + CD = AD AB 3 = CD 2 BC 1 and = AD 5 AB 3 = ⇒ CD 2 AC − BC 3 ⇒ = BD − BC 2
n ≥ 16
12
C
⇒ ⇒
P
∑
B
abc 4∆ (a ) (6) (6)
5= 4
⇒
5=
⇒
5=
12 + a 12 + a − 6 2 2 12 + a − 6 12 + a − a 2 2
36a (12 + a ) (a ) (a ) (12 − a ) 4 16 36 144 − a 2
188
KVPY Practice Set 2 Stream : SA
36 ⇒144 − a 2 = 5 ⇒
2
36 a 2 = 144 − 5
2
∴ Hence, two solution x = 1,
9 144 × 16 = 144 1 − = 25 25 ⇒
a2 =
3 x x − = 0 2 3 9 x = , x ≠ 0⇒x = ⇒ 2 4
144 × 16 48 = 25 5
11. (a) Let the second car overtakes in t hours. ∴Distance covered by first = Distance covered by second car 1 1 10t = 8 + 8 + + 8 + 2 2 2 t − 1 3 + 8 + + K + 8 + 2 2 1 ⇒ 10t = 8t + (1 + 2 + 3 + K + t − 1) 2 1 (t ) (t − 1) ⇒ 10t = 8t + 2 2 ⇒ t 2 − t = 8t ⇒ t − 1 = 8 ⇒ t = 9 ∴The second car overtake the first car in 9 h.
a2 =
9 4
15. (b) Given, 0 < a < b < c Here, a, b, c are distinct. ∴Three digits number formed by using a, b, c where digits are not repeated is 3! = 6 Sum of all the three digits number are 2! (a + b + c) (102 + 10 + 1) = 2(a + b + c) (100 + 10 + 1) = 222(a + b + c) Given, 222(a + b + c) = 1554 1554 a+ b+ c= =7 222 The possible digits whose sum seven are 1, 2, 4 ∴ b= 2
⇒ x2 + 2013 = y2 ⇒ y2 − x2 = 2013 ⇒ ( y + x) ( y − x) = 3 × 11 × 61 xy is minimum when y − x = 33 : y + x = 61 ∴ x = 14, y = 47 ∴Minimum value of xy = 14 × 47 = 658
16. (b) With respect to observer, there is no acceleration in the vertical velocity component. So, path of particle is a straight line as in option (b). 17. (c) F = − dU = − (2x + 3) dx For stable equilibrium, 3 F = 0⇒x = − 2 d 2U Also, =2 dx2 So, there is a minima ofU. i.e. system is in stable equilibrium.
13. (a) We have,
18. (c) By principle of thermometry,
12. (b) We have,
Edge of cube = 2 cm ∴ Total surface area of cube = 6 (side )2 = 6(2)2 = 24 cm 2 Length, breadth and height of cuboid are 1, 2 and 3 respectively Total surface area of cuboid = 2(lb + bh + hl) = 2(2 + 6 + 3) = 22 cm 2 Total surface area of both cube and cuboid is 24 + 22 = 46 which is less than 54 cm 2. ∴Both cube and cuboid can be painted.
14. (b) Given, (x)x = (x )
x x
= (x x ) x 3 2
=
Case I When base x = 1 Case II When base x ≠ 1 3x Then, x3 / 2 = 2
3x x2
25 − 0 X − (− 180° X) = 100 − 60° X − (− 180° X) X + 180° X 1 ⇒ = 120° X 4 120° X = 30° X X + 180° X = ⇒ 4 X = −180° X + 30° X ⇒ X = − 150° X
19. (d) Following conservation of mass number and atomic number, we have 9 4 12 1 4 Be + 2 He → 6 C + 0 n So, particle
1 0n
− F + µm2 g = − 1ms−2 m2
Now, from equation of motion, we have 1 s = ut + at 2 2 1 22 = 0 + (10 − (− 1)) t 2 2 ⇒ t = 2s 21. (a) Force on ball, F = v ∆m ∆t ∆m where, = flow rate of water. ∆t ∆m ⇒ = mg v ∆t ∆m mg m g = = ⇒ ∆t 2 gh 2 h
22. (c) For a cyclic process, 1→ 2→ 3→ 1 Σ∆U = 0 ⇒100 + c + (− 200) = 0 or c = 100 kJ
23. (d) From free body diagram,
a
= 20 N
mg
24. (c) Electrons are not effective in scattering αparticles because they are about 7000 times lighter than αparticles.
25. (b)
2h gh = g 1 − R
and
d gd = g 1 − R
⇒ ⇒
gh = gd ⇒ 2h = d h 1 = = 0.5 d 2
26. (b) India is in northern hemisphere and due to rotation of earth, radially rushing wind will tend to rotate in anticlockwise sense. 27. (d) Imagine earth rotating around sun and moon around earth. Path of moon
is a neutron. Sun
20. (d) m=0.2
m=0.5
A
B 22 m
Acceleration of blocks are F − f1 F − µm1 g a1 = = = 10 ms−2 m1 m1
F
F − mg = ma ⇒ F = m( g + a ) = 1 (10 + 10)
E
Path of earth
28. (d) Velocity increases with time and then reaches terminal velocity. Velocity remains constant after reaching terminal speed.
189
KVPY Practice Set 2 Stream : SA 29. (c) As density of ball
= 40 = 0.8 g cm −3 is less than water, it 50 tends to float. When ball is kept immersed, downthrust = weight of 50 cm3 of water = 50 g So, scale reading = 600 + 50 = 650 g
30. (b) Pcombination = 2(Plens ) + Pmirror 2 × 100 + 0 flens (in cm) 200 = − 12 / (15 . − 1) =
=
100 25 =− D 3 − 12
31. (c) CaCO3 → CaO + CO2 Molar mass of CaCO3 = 40 + 12 + 16 × 3 = 100 g Molar mass of CaO = 40 + 16 = 56 g Molar mass of CO2 = 12 + 16 × 2 = 44 g 100 g of CaCO3 produces 44 g of CO2 44 × 10 = 4.4 g ∴ 10 g of CaCO3 produces = 100 32. (c) According to Raoult’s law the vapour pressure of volatile component is directly proportional to its mole fraction. If the solution obeys Raoult’s law at all concentration its vapour pressure would vary linearly from zero to the vapour pressure of pure solvent. Thus, the correct plot will be (c). 33. (a) The ionic character is decided by Fajan’s rule. According to this rule, larger is size of cation, smaller the size of the anion and lesser is the charge on the cation or high, thus more will be the ionic character. As the anion and charge of n the cation in all the given compounds are same. So, the ionic character is only dependent on the size of cation. As the size of cation increases in the order Be2+ < Mg 2+ < Ca 2+ < Ba 2+ ∴ The ionic character will also increase in the same manner, i.e. BeCl 2 < MgCl 2 < CaCl 2 < BaCl 2 0
34. (d) 3Br2 + 6CO32− + 3H2O → 5Br− +5
+
B rO3−
+
6HCO3− −
In the reaction, Br2 is reduced to Br (oxidation number decreases from zero to − 1) and Br2 is oxidised to BrO3− (oxidation number increases from zero to +5 ). 35. (a) 1 2
6 5
H3 C
3 4
CH3
The IUPAC name of the above given compound is 3, 5dimethyl cyclohexene.
36. (d) As pentan2one (CH3 COCH2CH2CH3 ) and 3methyl butanone (CH3 COCH(CH3 )CH3 ) have similar molecular formula, but different carbon skeletons. Thus, they are chain isomers and will exhibit chain isomerism. 37. (a) The given value of n = 3 suggests that the shell is 3. For n = 1, l has 3 values, i.e. + 1, 0 and − 1hence there occur 3 orbitals in psubshell namely px , py and pz . Thus, the given values for n = 3, l = 1 and ml = 0 suggests that the orbital is 3py . Hence, the maximum number of orbitals that can be identified with given quantum number is only 1.
38. (a)
CH3CH==CHCH2CH3
O3
O CH3CH
direction where there are less number of moles (according to Lechatelier principle). 2SO2 ( g ) + O2 ( g ) 2SO3 ( g ) Hence, the reaction will more towards right.

41. (c) Given, solubility product of BaCl 2 , Ksp = 4 × 10−9 Let the solubility of BaCl 2 be S. BaCl 2 Ba 2+ + 2Cl − S S 2S Ksp = [Ba 2+ ] [Cl − ]2 = (S) (2S)2 = 4S 3

1
K 3 4 × 10−9 S = sp = 4 4
1/3
= 1 × 10−3 mol L−1
42. (a) Chlorobenzene on treatment with sodium in dry ether gives diphenyl. The reaction is known as Fittig reaction.
CHCH2CH3
O
—Cl + 2Na + Cl—
O ozonide H3O+/ZnO
Dry ether
+ NaCl
∆
CH3CHO + OHCCH2CH3
This reaction is known as ozonolysis reaction in which the addition of ozone molecule to alkene gives ozonide and then cleavage of ozonide by Zn—H2O to smaller molecules occurs. 39. (c) PH5 does not exist due to very less electronegativity difference between P and H. Hydrogen is slightly more electronegative than phosphorus, thus could not hold significantly the sharing electrons. On the other hand, BiCl5 does not exist due to inert pair effect. This is because on moving down the group, +5 oxidation state becomes less stable while +3 oxidation state become more stable due to inert pair effect. In SO2, pπdπ and pπ  pπ both types of bonds are present. SeF4 has sp3 dhybridisation whereas CH4 has sp3 hybridisation. Thus, they both have different geometry. I3+ has a bent shape due to the presence of 2 lone pairs on central I atom.
40. (a) Any change in the concentration, pressure and temperature of the reaction results in change in the direction of equilibrium. This change in the direction of equlibrium is governed by LeChatelier’s principle. On increasing pressure, volume decreases. The reaction will move in the
Diphenyl
43. (c) According to ideal gas equation, pV = nRT Given, Pressure = 360 torr =
360 atm 760
Volume = 2 .5 Temperature = 22 .5 + 273 = 295.5 K Weight of a gas = 16616 . pV n= ∴ RT W pV W WRT Also n = = ⇒ =M = M RT M pV Substituting the values, 1.6616 × 0.082 × 295.5 M= = 34.26 g 360 × 2.5 760 ≈ 34 g
44. (c) Isoelectronic species are those species which have same number of electrons. As all the given elements are isoelectronic with each other. Thus, the radii/size of isoelectronic species is inversely proportional to the atomic 1 number, i.e. size ∝ Z Thus, the correct order is S2− > Cl − > K+ > Ca 2+
190
KVPY Practice Set 2 Stream : SA
45. (c) The reaction of toluene with Cl 2 in the presence of FeCl3 gives predominantly o and p chlorotoluene. This reaction follows electrophilic substitution mechanism and Cl + act as as electrophile. CH3 CH3
Cl2, FeCl3 Cl+ (Chlorination) Toluene
Cl pchlorotoluene
CH3
+ Cl ochlorotoluene
46. (a) Phosphatidylcholine is a class of phospholipids that are a major component of biological membranes (i.e. nerve cell membrane). It functions in the production of brain chemical called acetylcholine used for nerve impulse transmission at the synapse.
47. (a) Graft rejection does not involve erythrocytes. Transplant or graft rejection occurs when transplanted tissue is rejected by the recipient’s immune system, which destroys the transplanted tissue. Rejection is an adaptive immune response via cellular immunity (mediated by killer Tcells), macrophages and polymorphonuclear leukocytes (i.e., neutrophils, eosinophils and basophils).
51. (c) The structures adopted by polypeptides can be divided into four levels of organisation, i.e., the primary, secondary, tertiary and quaternary structures. The secondary structure pertains to the coiling of the polypeptide chains into regular structure such as αhelices and βpleated sheets. 52. (c) At prophaseI, DNA replication has already occurred and the original amount of DNA has been doubled to x. At anaphaseI, the amount of DNA in the cell remains the same because no cytokinesis has occurred yet to separate the cytoplasm. 53. (b) If O2 is not available, pyruvic acid undergoes anaerobic respiration / fermentation, but under aerobic condition, the pyruvic acid enters into mitochondria and converted to acetyl CoA. Acetyl CoA functions as substrate entrant for Krebs’ cycle. So, it is a connecting link between glycolysis and Kreb’s cycle. 54. (b) Glucose is stored in the liver as glycogen. Glycogen can be converted to free glucose by the process of glycogenolysis, which involves the activation of a phosphorylase enzyme by the hormone glucagon. Glucagon is made by the pancreas and is released when the blood sugar levels fall. There release of glucose is a homeostatic function of liver that is controlled and monitored in the pancreas.
48. (b) Horseshoe crabs are marine and brackish water arthropods. They resemble crustaceans but belong to separate subphylum of the arthropods, i.e. Chelicerata. The entire body of the horseshoe crab is protected by a hard carapace.
55. (c) As the membrane potential is increased, sodium ions channels open, allowing the influx of Na + ions into the cell. The inward flow of sodium ions increases the concentration of positively charged cations in the cell and causes depolarisation, where the potential of the cell is higher than the cell’s resting potential.
49. (a) The first living being on the earth
56. (a) Ammonia is toxic waste product
were anaerobic or heterotrophic bacteria because the primordial atmosphere was virtually oxygenfree. These organisms must have degraded simple compounds present in the primordial oceans. They may have had RNA genomes and used RNA as biological catalysts.
which is converted into urea in the liver. This urea then enters the excretory system to get eliminated from the body. High levels of ammonia in blood is an indication of liver damage.
50. (b) Salmonella bacterium causes salmonellosis infection. The bacteria spread through human or animal faeces. Thus, the presence of Salmonella in tap water is due to contamination through human excreta. Salmonella outbreaks are commonly associated with eggs, meat and poultry, but these bacteria can also contaminate other foods such as fruits and vegetables.
59. (a) Centrosomes are made up of a pair of centrioles and other proteins. The centrosomes are important for cell division and produce microtubules that separate DNA into two new identical cells. 60. (a) The cerebellum is the part of hindbrain responsible for handeye coordination. It is responsible for maintaining equilibrium, transfer of information, fine adjustments to motor actions, coordinating eye movements, etc. Coordination and body balance, posture during walking, riding, standing, swimming, running are all maintained by the cerebellum.
61. (a) Given, ABC is an equilateral triangle. A
O
B
O′ r D′
P C
D
AB = BC = AC = 6 1 BD = BC = 3 2 AD = AB sin B AD = 6 × sin 60° 3 AD = 6 × =3 3 2 1 1 OD = AD = × 3 3 = 3 3 3 In ∆O ′ PO, ∠OO ′ P = 30° OP = OD − PD = 3 − r OO ′ = 3 + r 3+ r OP = ∴ sin 30° = OO ′ 3−r ⇒ ⇒ ⇒
57. (d) Mendel did not choose pod
1 = 2
3+ r 3−r
3 − r = 2 3 + 2r 1 r= 3
length. The seven contrasting traits he took were l Plant height l Flower position l Pod colour l Pod shape l Flower colour l Seed shape l Seed colour
62. (c) We have, x2 y3 = 6
58. (a) The levels of classification from
⇒
the broadest to the narrowest, i.e. in term of having highest members to the lowest members are kingdom, phylum, class, order, family, genus and species.
⇒ 3x + 4 y ≥ 10 ∴ Minimum value of 3x + 4 y is 10.
3x 3x 4 y 4 y 4 y 1/5 + + + + 2 3 2 2 3 3 3 ≥ 9x × 64 y 4 5 27 [∴ AM ≥ GM ] 16 × 6 3x + 4 y ≥ 5 3
1/5
[x2 y3 = 6]
191
KVPY Practice Set 2 Stream : SA 63. (c) We have three different kinds of mangoes and we can select 25 mangoes in all. Hence, we select 0 or 1 or 2 or 3 ... mangoes from each kind of mangoes. Let x1 , x2 and x3 be different kinds of mangoes. ∴ x1 + x2 + x3 = 25 ⇒ x1 , x2 , x3 ≥ 0 ∴Total number of selection = 25 + 3 − 1C3 − 1 = 27C2 27 × 26 = 351 = 1× 2
64. (d) We have, (1 − y)m (1 + y)n m(m − 1) 2 y + K = 1 − my + 2 n ( n − ) 1 1 + ny + y2 + K 2 ⇒ (1 − y)m (1 + y)n = 1 + (n − m) y n (n − 1) m(m − 1) + + − nm y2 + K 2 2 a1 = n − m = 10
Here, a2 = ⇒ a2 =
...(i)
n (n − 1) m(m − 1) + − nm = 10 2 2 n 2 − n + m2 − m − 2mn = 10 2
⇒ (n − m)2 − (n + m) = 20 From Eqs. (i) and (ii), we get n + m = 80 From Eqs. (i) and (ii), we get n = 45, m = 35 ∴ (m, n ) = (35, 45)
...(ii) ...(iii)
66. (c) Present temperature, T1 =
+ 20 × (500) 2 + 20 × ( 800) 2 + 20 × (1000) 2 ) 3 × 138 . × 10 −23
= =
10 −1
3 × (10 + 120 + 500 + 1280 + 2000) 3
3910 = 130.33 ≈ 130 K 30
In few years, all the molecules with speed > 900 ms −1 will left the atmosphere. So, then temperature will be . × 10−28 × (10 × (100)2 + 30 × (200)2 138 T2 =
+20 × (500)2 + 20 × (800)2 )
. × 10−23 3 × 138 10−5 × 104 × (10 + 30 × 4 + 20 × 25 + 20 × 64) = 3 10−1 × (10 + 120 + 500 + 1280) = 3 1910 = = 63.66 K 30
ruler. Forces on ruler are as shown below. 2 L/ θ
θ
–θ M
xL 2xL
N A
For convex lens, image is at focal distance 6 cm from the lens. For concave lens, u = + 2 cm, f = − 2.5 cm 1 1 1 1 1 = + = + ⇒ v f u −2.5 2 ⇒ v = + 10 cm Linear magnification, produced by v 10 diverging lens is m = = =5 2 u So, diverging lens increases the magnification five times.
70. (c) ⇒ ⇒
I = nevd A I V /R vd = = neA neA V vd = ρlne
71. (d) Given, specific heat of water
L
∠A = 90°, the angle bisector of B and C meet at P The distance from P to hypotenuse is 4 2 PQ = 4 2 ∴ Here, PQ is the radius of incircle of ∆ABC. ∴PQ = PM = PN = radii of incircle of ∆ABC ∴ AP 2 = PM 2 + AM 2 AP 2 = (4 2 )2 + (4 2 )2 [Q AM = PN ] AP 2 = 32 + 32 AP = 64 = 8
A′′
67. (a) Let x = mass per unit length of
90 C
A′
So, drift speed is independent of area of conductor. Hence, the ratio of vd$ /vd$ is 1.
4 √2 4 √2 4 √2
B′′
Reduction in temperature is nearly, ∆T = T1 − T2 = 130 − 63.66 = 66.34 K ≈ 70 K
ABC is a right angle triangle.
P
B′
× 64 + 20 × 100)
=
P
Q
Rays from object
10 −5 × 10 4 (10 + 30 × 4 + 20 × 25 + 20
65. (a) We have, B
69. (c)
138 . × 10 −28 × (10 × (100) 2 + 30 × (200) 2
Equating moments about P, L (x ⋅ L) sin (90 − θ) = L (2xL) sin θ 2 1 sin θ = tan θ = ⇒ 4 cosθ
68. (a) 1 kg of water expands from 1000 cm3 to 1.68 m3 . ∴ ∆V = 168 . − 0.001 ≈ 168 . m3 So, work done in expanding against pressure is ∆W = p∆V = 101 × 10 3 × 168 . = 169 kJ 169 ∆W Now, = = 0.0748 1 × 2260 mLV
2
1
= 4.184 g −1 ° C−1 Heat capacity of calorimeter = 783 J°C −1 Mass of water = 254 g ∆T = 26.01 − 23.73 = 2 .28°C qbomb = C∆T = 783 × 2 .28 J = 1785.24 J qwater = m × specific heat × ∆T = 254 × 4184 . × 2 .28 = 2423.04 J Heat absorbed = 1785 .24 J + 2423.04 J = 4208.28 J ≈ 4.21 kJ
72. (b) The hybridisation of any molecule can be calculated using formula 1 X = (valence electrons number of atoms 2 monoatomic ± anion/cation) The shape and hybridisation of given molecules are as follows Molecule Shape
Hybridisation
NF3
Pyramidal
sp 3
BF3
Triangular planar
sp 2
BF4−
Tetrahedral
sp 3
NH4+
Tetrahedral
sp 3
192
KVPY Practice Set 2 Stream : SA
Molecule Shape
Hybridisation
BCl 3
Triangular planar
BrCl 3
Tshaped
sp
2
sp 3d
CH CH 3CH 2OH H3 C — C Br → SN 1 CH3
Pyramidal
sp
NO −3
Triangular planar
sp 2
As BF4− and NH+4 have same shape and hybridisation. Thus, they are isostructural pair.
73. (a) Geometrical isomers are those isomers, which have same molecular formula, but different spatial arrangement of atoms about the double bond. In geometrical isomers, both the carbon atoms of a double bond should contain different substituents. xH Hx
As in the given compound is 3°alkyl halide, so it can either go SN 1or E1 reaction, which is dependent on the nutrophile. As CH3 CH2O– Na + is a strong Nu s, so it will prefer to undergo E1 reaction whereas CH3 CH2OH is a weak Nu s which undergoes SN 2.
75. (b) Number of π and σ bands in
As both the substituents are different hence, they will show geometrical isomerism. y H3C CH3 x
Structure σ π σ C σ
(b)
No. of σ and π bonds σ bonds = 4 π bonds = 1
O—H σ σ bonds = 4
O
σ π
σ Xe σ O πσ π π O
CH3 x
π bonds = 4
O
It does not show geometrical isomerism. Hx (c) x y H
O
–O
(b) y CH3
Hx
It does not show geometrical isomerism. Hx (d)
(c)
2π σ
(d)
σ
2π
σ bonds = 3
σ
π bonds = 4
N ≡≡ C C ≡≡ N
σ bonds = 6 H σ 2π 2π π bonds = 4 σ σ C ≡≡ N N ≡≡ C C σ σ σ H
76. (c) The graph (c) is correct as, it
Hx
It does not show geometrical isomerism. CH3 CH 3CH 2O− Na + 74. (c) CH3 C == CH2 ← E1
I BI O
IA IO
IA IO
IOIO
IB IB
×
(ii) IA IB
I BI O
I AI B
IAI A
IB I O
I BI B
×
(iii) I AIB
I AI B
I AIB
IAI A
I AIB I BI O
×
(iv)
given species are as follows :
(a)
CH3 y
I AI O
( Major) ( A)
(a) y H3C
I BI O
×
(i)
CH3 CH3 COCH2CH3 CH3
3
NH3
I AI O
shows the highest biochemical activity in the lysosomes. These are membrane bound organelles which contain enzymes that degrade polymers into their monomeric subunits, i.e. hydrolytic enzymes.
77. (a) There are four possible mating crosses which can occur with persons of blood group A and blood group B.
I AIB
I AIO
I AI B
I AIO
Thus, in all four possibilities for F1 generation can be IA IB , IA IO, IBIO , IOIO but never IA IA or AA.
78. (b) Only the electron transport system produces ATP and reduced NADP in the plant. Without these compounds, the Calvin cycle cannot proceed and carbonfixation cannot occur and their is no respiratory substrate available for respiration. 79. (a) Immovable / fixed / fibrous joints are present between the skull bones. So, between parietal bone and the temporal bone of the skull are joined by fibrous joint. Other statements can be corrected as First cervical vertebra is atlas not axis. The 11th and 12th pairs of ribs are called floating ribs. Glenoid cavity is located at the end of scapula close to coracoid process.
80. (c) The triplet codon is made up of 3 nucleotide bases and is located at the centre of the middle loop of tRNA molecule and base pairs with the complementary bases on an mRNA molecule during protein synthesis.
193
KVPY Practice Set 3 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
PRACTICE SET 3 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS 1. Let E (n ) denote the sum of the even digits of n. For example E(1243) = 2 + 4 = 6, then the value of E (1) + E (2) + E (3) + K + E (100) is equal to (a) 200
(b) 300
(c) 400
(d) 500
2. The greatest possible perimeter of right angle
(a) is at least 30 (b) is at least 20 (c) is exactly 25 (d) cannot be determined by the data
6. Let a, b and c such that a + b + c = 0 and is defined as P=
a2
triangle with integer side length if one of the sides has length 12 is
2a + bc is equal to
(a) 80
(a) 1
(b) 84
(c) 72
(d) 82
3. In a party, each man danced with exactly four women and each woman danced with exactly three men. Nine men attended the party, then number of woman attended the party is (a) 12
(b) 9
(c) 6
(d) 8
4. If 3 + 2 = 985 and 3 − 2 = 473, then the value of xy x
y
x
y
is (a) 36
(b) 72
(c) 48
(d) 54
5. A certain school has 300 students. Every student reads 5 newspapers and every newspaper is read by 60 students. Then, the number of newspaper
(c)
2
+
b2 2b + ac 2
+
c2 2c + ab 2
(b)
1 4
, then the value of P
1 2
(d) 2
7. In a rectangle ABCD, AB = 8 and BC = 20, let P be a point on AD such that ∠BPC = 90°. If r1 , r2 and r3 are radii of the incircles of ∆ APB, BPC and CPD respectively, then the value of r1 + r2 + r3 is equal to (a) 6
(b) 8
(c) 10
(d) 12
8. If sin θ + cos θ = 3, then the value of tan θ + cot θ is (a) 1 (c) 2
(b) 2 (d) None of these
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KVPY Practice Set 3 Stream : SA
9. Numbers 1, 2, 3, ... 100 are written down each of the cards A, B and C. One number is selected at random from each of the cards. The probability that the numbers so selected can be the measures (in cm) of three sides of the right angled triangles no two of which are similar is (a)
4 1003
(b)
3
(c)
503
3! 1003
(d) None of these
10. In a triangle with integer side length, one side is three times as long as a second side and the length of the third side is 17. What is the greatest possible perimeter of the triangle? (a) 46
(b) 47
(c) 48
(d) 49
11. One morning, each member of Kanchan’s family drank 8 ounce mixture of coffee and milk. The amount of coffee and milk varied from cup to cup, but 1 were never zero. Kanchan drank th of the total 7 2 amount of milk and th of the total amount of 17 coffee, then the number of people are there in Kanchan’s family are (a) 8
(b) 9
(c) 7
(d) 17
12. Let ABC be a triangle with ∠ABC = 90°. Let P and Q are midpoint of legs AB and BC, respectively. Suppose that AQ = 19 and PC = 22, then length of AC is equal to (a) 24
(b) 25
(c) 26
(d) 30
13. Number of natural numbers n between 1 and 2019 (both inclusive) is (a) 0
8n an integer is 9999 − n
(b) 1
(c) 2
(d) 3
ABCD passes through the vertex B after getting reflected by BC, CD and DA in that order. If θ is the angle of the initial position of the ray with AB, then sin θ equals 2 13
(b)
3 13
(c)
3 5
(d)
4 5
15. Let m be the number of ways in which two couples can be seated on 4 chairs in a row, so that no wife is next to her husband and n be the number of ways in which they can be seated in 4 chairs in a circle. In the other case rotation are considered different configurations. Then, (a) m = n
(b) m = 2n
(c) m = 4n
17. Potential energy of a spring when stretched through a distance x is 10 J. Ratio of work done for every additional distance is (a) 1 : 1 : 1 : .... (c) 1 : 3 : 5 : ....
(b) 1 : 2 : 4 : ..... (d) 1 : 4 : 9 : 16 : ...
18. At what temperature, the celsius and farenheit scale give the same temperature value? (b) − 10°F (d) − 20°C
(a) 0°C (c) − 40°C
19. In given nuclear transformation, 92 U
238
−
α → B Th A β →
C D Pa
E →
92U
234
A, B, C , D and E are (a) A = 234, B = 90, C = 234, D = 91, E = β (b)A = 234, B = 90, C = 238, D = 94, E = α (c) A = 238, B = 93, C = 234, D = 91, E = β (d)A = 234, B = 90, C = 234, D = 93, E = α
20. A particle is subjected to two simple harmonic motions π (x = 2 sin ωt) cm and x = 2 sin ω t + cm 3 time t is in seconds. Maximum speed of the particle, rad is if ω = 1 s (a) 6
14. A ray of light originating at the vertex A of a square
(a)
The correct option is (a) HI > HII > HIII (b) HI < HII < HIII (c) HI = HII < HIII (d) data insufficient to conclude
(c) 2
cm
(b) 3
s2 cm
cm
s2 1 cm (d) 2 s2
s2
21. A man crosses a 320 m wide river perpendicular to the current in 4 min. If in still water, he can swim 5 with a speed times that of the current, then the 3 speed of the current (in m/min) is (a) 30
(b) 40
(c) 50
(d) 60
22. Consider a 16 cm × 8 cm uniform rectangular sheet with its sides parallel to axes and its centre at origin.
(d) m = 8n
(0,0)
PHYSICS 16. A sample of pure ice is taken and following are recorded. I. Heat to melt the ice. II. Heat to warm ice cold water to 100°C. III. Heat to vaporize the water at 100°C.
If exactly one quarter of this sheet is removed, coordinates of centre of mass of remaining sheet are (a) (4 / 3, 2 / 3) (c) (4 / 3, 0)
(b) (0, 2 / 3) −4 −2 (d) , 3 3
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KVPY Practice Set 3 Stream : SA U
23. For four processes A, B, C and D, loge p versus loge V graph are given below. ln p
(a)
F
F
(b)
B
C
U
D
r0
r0
r
r
A
(d)
(c)
ln V F
Isothermal process is (a) A
(b) B
(c) C
U
r0
(d) D
F
r0 r
24. In given set up,
U
focal length of mirror = 20 cm, focal length of lens = 15 cm and separation of mirror and lens = 40 cm. A point source S of light ‘S’ is placed on principal axis at distance d from lens.
29. A metal sphere is held suspended along a wall as shown below.
If the final beam comes out parallel to the principal axis, then value of d is
12 cm
f=20 cm
f=15 cm
r=5 cm
S
If string can break at a pull of 15 N, then maximum density of material of sphere can be (a) 4 cm
(b) 8 cm
(a) 44 kg m −3 (b) 64 kg m −3 (c) 54 kg m −3 (d) 74 kg m −3
40 cm
(c) 12 cm
(d) 16 cm
= K , where p = pressure, V = volume and K = constant. The dimensions of constant K are
25. A gas satisfies the relation pV
5/ 3
(a) [ML4 T −2 ] (b) [ML2 T −2 ] (c) [M0 L0 T 0 ] (d) [MLT −2 ]
26. When temperature of a semiconductor is raised, then choose the correct option. (a) None of electron jump to higher energy level (b) All electrons likely to jump at higher energy levels (c) Electrons whose energies are close to fermi energy are likely to jump to higher energy levels (d) Electrons having lesser energy than fermi energy are likely to jump to higher energy level
27. An incense stick is lighted in a closed room in which there is no flow of air. Then, choose the correct option given below. (a) Flow of smoke is initially turbulent, then laminar (b) Flow of smoke is initially laminar, then turbulent (c) Flow of smoke is turbulent only (d) Flow of smoke is laminar only
28. Which of these graphs correctly shows potential energy and force between two atoms in a diatomic molecule? (U = potential energy, F = force, r = distance)
30. An object at infinity forms an image of size 2 cm by a convex lens of focal length 30 cm. Now, a concave lens of focal length 20 cm is placed between the convex lens and image at a distance of 26 cm from convex lens. Image size now will be (a) 1.25 cm
(b) 2.5 cm
(c) 1.05 cm
(d) 2 cm
CHEMISTRY 31. If 500 mL of a 5M solution is diluted to 1500 mL. What will be the molarity of the solution obtained? (a) 1.5 M
(b) 1.66 M
(c) 0.017 M
(d) 1.59 M
32. A plot of volume (V ) versus temperature (T ) for a gas at constant pressure is a straight line passing through the origin. p1 Volume (mL)
d
p2 p3 p4 Temperature (K)
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KVPY Practice Set 3 Stream : SA
The plot of different values of pressure are shown in figure. Which of the following orders of pressure is correct? (a) p1 > p2 > p3 > p4 (c) p1 < p2 < p3 < p4
44. The correct IUPAC name of the following compound
(b) p1 = p2 = p3 = p4 (d) p1 < p2 = p3 < p4
is
33. A substance X gives brick red flame and breaks down on heating to give oxygen and a brown gas. The substance X is (a) magnesium nitrate (c) barium nitrate
(b) calcium nitrate (d) strontium nitrate
34. The major product obtained when 1 butyne reacts
(a) 3(1ethyl propyl) hex1ene (b) 4ethyl3propyl hex1ene (c) 3ethyl4ethenyl heptane (d) 3ethyl4propyl hex5ene
45. The period number in the long form of the periodic
with excess HBr is
table is equal to
(a) 2, 2dibromobutane (b) 2bromobutane (c) 1, 1, 2, 2tetrabromobutane (d) 1, 2dibromobutene
(a) magnetic quantum number of any element of the period (b) atomic number of any element of the period (c) maximum principal quantum number of any element of the period (d) maximum azimuthal number of any element of the period
35. Which of the following sulphides when heated strongly in air gives the corresponding metal? (a) Cu 2S
(b) CuS
(c) Fe2S3
(d) HgS
36. Sulphur in + 3 oxidation state is present in (a) dithionous acid (c) dithionic acid
(b) sulphurous acid (d) pyrosulphuric acid
37. The correct order of boiling points of given compounds is nbutylamine (I), diethyl amine (II), N, Ndimethylethylamine (III). (a) III < II < I (c) III < I < II
(b) I < II < III (d) II < I < III
38. The value of Planck’s constant is 663 . × 10−34 Js. The −1
velocity of light is 3 × 10 ms . Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 8 × 10−15 s−1? 8
(a) 2 × 10−25 (c) 4 × 101
(b) 3 × 107 (d) 5 × 10−18
39. The correct order of electron affinities of N, O, S and Cl is (a) N < O < S < Cl (c) O ≈ Cl < N ≈ S
(b) O < N < Cl < S (d) O < S < Cl < N
40. The number of lone pairs on central metal atom Xe in XeF2 , XeF4 and XeF6, respectively are (a) 2, 3, 1
(b) 1, 2, 3
(c) 4, 1, 2
(d) 3, 2, 1
41. When aqueous solution of benzene diazonium chloride is boiled, the product obtained is (a) C6 H5 CH2OH (c) C6 H5 COOH
(b) C6 H6 + N2 (d) C6 H5 OH
42. The set of quantum number for 19th electron of chromium (Z = 24) is
1 2 1 (c) 3, 2, 2, + 2
(a) 4, 0, 0, +
(b) 4, − 1, 1, − 1, + (d) 3, 2, − 2, +
1 2
1 2
43. Among the given carbocations, the most stable carbocation is (a) methyl
(b) allyl
(c) benzyl
(d) vinyl
BIOLOGY 46. Which of the following biomolecules is common to respirationmediated breakdown of fats, carbohydrates and proteins? (a) Glucose6phosphate (b) Fructose1,6bisphosphate (c) Pyruvic acid (d) Acetyl CoA
47. You are given a tissue with its potential for differentiation in an artificial culture. Which of the following pairs of hormones would you add to the medium to secure shoots as well as roots? (a) IAA and gibberellin (b) Auxin and cytokinin (c) Auxin and abscisic acid (d) Gibberellin and abscisic acid
48. The partial pressure of oxygen in the alveoli of the lungs is (a) equal to that in the blood (b) more than that in the blood (c) less than that in the blood (d) less than that of carbon dioxide
49. When cell has stalled DNA replication fork, which checkpoint should be predominantly activated? (a) G1 / S (b) G2 / M (c) M (d) Both G2 / M and M
50. Which of the following is the least likely to be involved in stabilising the threedimensional folding of most proteins? (a) Hydrogen bonds (b) Electrostatic interaction (c) Hydrophobic interaction (d) Ester bonds
197
KVPY Practice Set 3 Stream : SA 51. Name a peptide hormone which acts mainly on hepatocytes, adipocytes and enhances cellular glucose uptake and utilisation. (a) Insulin (c) Secretin
(b) Glucagon (d) Gastrin
homozygous for normal colour vision, the probability of their son being colourblind is (b) 0.5 (d) 1
53. One of the major components of cell wall of most fungi is (a) peptidoglycan (c) hemicellulose
56. Lack of relaxation between successive stimuli in sustained muscle contraction is known as
52. If a colourblind man marries with a woman who is (a) 0 (c) 0.75
(c) release bicarbonate ions by the liver (d) reduce the rate of heartbeat
(a) fatigue (c) tonus
(b) tetanus (d) spasm
57. Which of the following guards the opening of hepatopancreatic duct into the duodenum? (a) Ileocaecal valve (c) Sphincter of Oddi
(b) Pyloric sphincter (d) Semilunar valve
58. Which one of the following cell organelles is enclosed by a single membrane?
(b) cellulose (d) chitin
54. A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F1plants were selfed, the resulting genotypes were in the ratio of (a) 1 : 2 : 1 :: Tall heterozygous : Tall homozygous : Dwarf (b) 3 : 1 :: Tall : Dwarf (c) 3 : 1 :: Dwarf : Tall (d) 1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf
55. Reduction in pH of blood will
(a) Chloroplasts (c) Nuclei
(b) Lysosomes (d) Mitochondria
59. Which of the following features is not present in the phylum Arthropoda? (a) Metameric segmentation (b) Parapodia (c) Jointed appendages (d) Chitinous exoskeleton
60. Water soluble pigments found in plant cell vacuoles are
(a) reduce the blood supply to the brain (b) decrease the affinity of haemoglobin with oxygen
(a) chlorophylls (c) anthocyanins
(b) carotenoids (d) xanthophylls
PARTII (2 Marks Questions) MATHEMATICS
65. In a ∆ ABC, let I denotes the incenter. Let the line AI,
61. Let S be a set of real numbers with mean m. If the
means of set S ∪ {15} and S ∪ {15, 1} are m + 2 and m + 1 respectively. Then, number of elements S has (a) 4
(b) 5
(c) 6
(d) 7
62. For natural numbers x and y, let (x, y) denote the greatest common divisor of x and y. The pairs of natural number x and y with x ≤ y satisfy the equation xy = x + y + (x, y) is (a) 2
(b) 3
(c) 4
(d) 5
63. In a ∆ ABC, X and Y are points on the segment AB and AC respectively, such that AX : XB = 1 : 2 and AY : YC = 2 : 1. If the area of ∆ AXY is 10, then the area of ∆ ABC is
(a) 30
(b) 45
(c) 60
(d) 27
64. Let x1 , x2 , x3 , K , x2019 be the real numbers different from 1, such that x1 + x2 + x3 + K + x2019 = 1 and x1 x2 x2019 + +K+ = 1, then the value of 1 − x1 1 − x2 1 − x2019 2 x12 x22 x2019 is equal to + +K+ 1 − x1 1 − x2 1 − x2019
(a) 0
(b) 1
(c) 2019
(d) None of these
BI and CI intersects the incircle at P, Q and R, respectively. If ∠BAC = 40°, then the value of ∠ QPR in degree is (a) 50° (c) 55°
(b) 65° (d) 60°
PHYSICS c + t3
, where t is 66. Power is given by P = a + bt 2 + d time. Then, I. [a ] = [ML2T−3 ] II. [b] = [ML2T−5 ] III. [c] = [T3 ] IV. [d ] = [ML− 2T−6 ] Which of the above statements are correct? (a) Statements I and IV are correct (b) Statements I and III are correct (c) Statements I, II and III are correct (d) All statements are correct
198
KVPY Practice Set 3 Stream : SA
67. A rolling sphere collides with a cube of equal mass. Surface is frictionless. Radius of sphere is 1 cm and its initial angular speed is 1 radian per second. ω 2 cm
r=1 cm
If side of cube is 2 cm and collision is elastic, then after collision, (a) ωsphere = 0 and vcube = 0.01 ms−1 rad (b) ωsphere = 1 and vcube = 0.01 ms−1 s (c) ωsphere = 0 and vcube = 0 rad and vcube = 1ms−1 (d) ωsphere = 1 s
U =
− 2x x +4 2
Then, angular frequency of small oscillations of the body about the position of stable equilibrium is (a) 5 rad s−1 (c) π rad s−1
(b) 10 rad s−1 (d) 25 rad s−1
CHEMISTRY 71. The heat of formation of C12H22O11 (s), CO2 ( g) and
. and −683 . kcal/mol, H2O(l) are −530, − 943 respectively. The amount of C12H22O11 required to supply 2700 kcal of energy is
68. A current of 1 mA enters the network of resistors as shown below.
(a) 382.70 g (c) 463.9 g
(b) 832.74 g (d) 682.6 g
72. The following alcohol after treatment with acid gives compound A. Ozonolysis of A gives nonan 2, 8 dione. The compound A is
2kΩ
CHOHMe A
1kΩ
1kΩ
1kΩ
B
H+
A
Me
2kΩ
Now, consider the following statements: 2 I. Current through 2 kΩ resistor is mA. 5
(a)
(b)
1 mA. 5 1 III. Current through middle 1 kΩ resistor is mA. 5 3 IV. Current through middle 1 kΩ resistor is mA. 5 Which of the above statements are correct?
(c)
(d)
II. Current through lower 2 kΩ resistor is
(a) Only statement I is correct (b) Statements II and III are correct (c) Statements I and III are correct (d) Statements III and IV are correct
73. An electric current is passed through silver nitrate solution using silver electrodes. 10.79 g of silver was found to be deposited on the cathode. If the same amount of electricity is passed through copper sulphate solution using copper electrodes, the weight of copper deposited on the cathode is (a) 1.6 g (c) 3.2 g
(b) 2.3 g (d) 6.4 g
74. Suppose 10−17 J of light energy is needed by the
69. A steel ball travels through a hollow Utube with separation in limbs 1 m, as shown below.
interior of human eye to see on object. Calculate the number of photons of green light (λ = 550 nm) needed to generate this minimum amount of energy. (a) 26
1m
(c) 28
(d) 29
75. The Gibbs’ free energy change, ∆G°, for the following reaction is 63.3kJ
v
Radius of Utube is slightly greater than ball such that ball travels through it and leaves from other end with same uniform speed of 3 ms−1. Mass of steel ball is 0.02 kg. What is the approximate reading on spring balance attached to the tube? (a) 0.1 N
(b) 27
(b) 0.2 N
(c) 0.5 N
(d) 0 N
70. A body of mass m (= 5 g), is moving in one dimension under influence of a conservative force. Potential energy of the body is given by
Ag 2CO3 (s) º
2Ag + (aq) + CO32− (aq)
The Ksp of Ag 2CO3 (s) in water at 25°C is closest to
(a) 3.2 × 10−26 (c) 2.9 × 10−3
(b) 8 × 10−12 (d) 7.9 × 10−2
BIOLOGY 76. Impulses travel very rapidly along nerves to the leg muscles of a mammal. Which fact accounts for the speed at which they travel?
199
(c)
Substrate concentration
77. In a certain plant, yellow fruit colour (Y) is dominant to green (y) and round shape (R) is dominant to oval (r). The two genes involved are located on different chromosomes. Which of the above will result when plant YyRr is selfpollinated?
molecular weight of 110. The DNA strand coding for a polypeptide chain of molecular weight 20,000 has a length of (a) 182 nucleotides (b) 252 nucleotides (c) 540 nucleotides (d) 760 nucleotides
78. Which graph shows the expected relationship
80. Which of the following is a correct statement? (a) Salvinia, Ginkgo and Pinus all are gymnosperms (b) Sequoia is one of the tallest trees (c) The leaves of gymnosperms are not welladapted to extremes of climate (d) Gymnosperms are both homosporous and heterosporous
Rate of reaction
Rate of reaction
between enzyme activity and substrate concentration?
(b)
Substrate concentration
79. Assume that the average amino acid residue have a
(a) 9 : 3 : 3 : 1 ratio of phenotypes only (b) 9 : 3 : 3 : 1 ratio of genotypes only (c) 1 : 1 : 1 : 1 ratio of phenotypes only (d) 1 : 1 : 1 : 1 ratio of phenotypes and genotypes
(a)
(d)
Rate of reaction
(a) A nerve impulse is an allornothing phenomenon (b) The nerves contain myelinated fibres (c) There is high concentration of Na + ions inside the axons (d) There is a potential difference across the axon membranes
Rate of reaction
KVPY Practice Set 3 Stream : SA
Substrate concentration
Substrate concentration
Answers PARTI 1 11 21 31 41 51
(c) (a) (d) (b) (d) (a)
(b)
2 12 22 32 42 52
(a)
62 72
(d)
(c) (d) (c) (a)
(a)
3 13 23 33 43 53
(d)
63 73
(c)
(b) (c) (b) (c)
(c)
4 14 24 34 44 54
(d)
64 74
(c)
(a) (c) (a) (b)
(c)
5 15 25 35 45 55
(b)
65 75
(b)
(c) (a) (d) (c)
(a)
6 16 26 36 46 56
(b)
66 76
(b)
(b) (c) (a) (d)
(b)
7 17 27 37 47 57
(c)
67 77
(a)
(c) (b) (a) (b)
(d)
8 18 28 38 48 58
(b)
68 78
(b)
(c) (d) (c) (b)
(d)
9 19 29 39 49 59
(b)
69 79
(c)
(a) (a) (a) (a)
(d)
10 20 30 40 50 60
(c)
70 80
(b)
(a) (b) (d) (d)
PARTII 61 71
(a) (d)
(b)
(b)
(a)
(c)
(c)
(b)
(b)
(b)
(a)
200
KVPY Practice Set 3 Stream : SA
Solutions 1. (c) E (1) + E (2) + E (3) + K + E (100) = Sum of all even digits from 1 to 100 = Sum of all even digits in [01 + 02 + 03 + K + 98 + 99 + 100] = 0 × 20 + 2 × 20 + 4 × 20 + 6 × 20 + 8 × 20 [Q there are 2 × 100 = 200 digits and 200 each digit appears = 20 times] 10 = (2 + 4 + 6 + 8) × 20 = 20 × 20 = 400
2. (b) Let the other sides of right angle triangle be x and y. P x
R
y
Q
12
∴ x = 12 + y2 2 ⇒ x − y2 = 144 ⇒ (x + y) (x − y) = 72 × 2 x and y are integer. ∴Maximum value of x + y = 72 ∴Maximum perimeter of triangle = 12 + x + y = 12 + 72 = 84 2
2
5. (c) We have, Total number of students = 300 One student read = 5 newspapers Number of newspaper read by 300 students = 5 × 300 = 1500 Number of different newspaper 1500 = = 25 60
6. (a) Given,a + b + c = 0 ∴
a3 + b3 + c3 = 3abc a2 b2 c2 Now, P = + + 2a 2 + bc 2b2 + ac 2c2 + ab
Let a = 1, b = − 1, c = 0 1 1 P= + + 0 ∴ 2+ 0 2+ 0 1 1 ⇒ P = + =1 2 2
7. (b) Given, ABCD is a rectangle.
4. (c) Given, ...(i) 3x + 2y = 985 ...(ii) 3x − 2y = 473 On adding Eqs. (i) and (ii), we get 2 ⋅ 3x = 1458 3x = 729 3x = 36 ⇒ x=6 On subtracting Eq. (ii) from Eq. (i), we get 2 ⋅ 2y = 512 2y = 256 2y = 28 ⇒ y=8 ∴ xy = 6 × 8 = 48
4
A C1 8
3. (a) There are 9 men and each man danced with 4 women. ⇒ Number of dancing pairs = 9 × 4 = 36 Now, let number of women = x ∴Each woman danced with 3 men. ∴Number of dancing pairs = 3x ∴ 3x = 36, x = 12 Hence, 12 women attended the party.
AB = 8, BC = 20 AP = x PD = 12 − x
Let ∴
16
P
r3
r1 r2
D C3
r2 C2
B
12
C
In ∆APB and ∆DPC, ∆PBA ~ ∆CPD AB AP = ∴ PD CD 8 x = ⇒ 12 − x 8 ⇒ ⇒
x2 − 20x + 64 = 0 (x − 16) (x − 4) = 0 x = 4, 16 In ∆APB, 1 × AB × AP ∆ 2 r1 = = 1 s (AB + PB + BP ) 2 8× 4 r1 = 8+ 4+ 4 5 [Q BP 2 = AP 2 + BA 2 BP = 16 + 64 = 4 5] r1 = 12 − 4 5 Similarly, r2 = 6 5 − 10
r3 = 6 − 2 5 ∴r1 + r2 + r3 = 12 − 4 5 + 6 5 − 10 + 6 − 2 5 =8
8. (d) Given, sin θ + cos θ = 3 Squaring both sides, sin 2 θ + cos2 θ + 2 sin θ cos θ = 3 ⇒ 1 + 2 sin θ cos θ = 3 ⇒ sin 2 θ = 2 ∴Maximum value of sin θ = 1 Hence, sin θ + cosθ = 3 not possible.
9. (d) Given, Number 1, 2, 3, ... 100 are written on card A , B and C. ∴ Total number of outcomes = 100 × 100 × 100 = 1003 Numbers are selected such that formal three sides of right angle triangle and triangle are not similar. For a right angle triangle (2n + 1)2 + (2n 2 + 2n )2 = (2n 2 + 2n + 1)2 , n ∈N for n = 1, 2, 3, 4, 5, 6. If n > 6 then length of longest side is greater than 100. ∴ n (E ) = 6 × 3! 6 × 3! P (∈) = (100)3 6× 6 9 = = 3 2 × (50)3 2(50)3
10. (d) Given, One side of triangle is three times the second side and third side is 17. Let the sides of triangle are x, 3x, 17. We know, in triangle sum of two sides is greater than third side. ⇒ x + 3x > 17 17 x> ⇒ 4 and ∴ ∴
x + 17 > 3x 17 x< 2 17 17 < x< 4 2 4.25 < x < 8.5
Since, sides of triangle are integer. ∴
x = 5, 6, 7, 8
∴Maximum value of x = 8 ∴Sides of triangle are 8, 24, 17 ∴Perimeter = 8 + 24 + 17 = 49
201
KVPY Practice Set 3 Stream : SA 11. (a) Let the total milk is 7M ounce and total coffee is 17C ounce. The ratio of total milk and coffee drank by Kanchan’s must be integer. 7M + 17C be an integer (which is the ∴ M + 2C Total number of people in Kanchan’s family) 7M + 17C 3C = 7+ M + 2C M + 2C 3C 3 ∴ < 0< M + 2C 2 3C =1 M + 2C 3C 3 is integer less than ] [Q M + 2C 2 C= M ∴Total number of people in Kanchan’s 7M + 17M family =8 M + 2M
12. (c) Given, ABC is a right angled triangle ∠ABC = 90° P and Q are midpoints of sides AB and BC respectively. C
4 4 (361 + 484) = × 845 5 5 ⇒ AC 2 = 4 × 169 AC = 4 × 169 = 26 ⇒ 8n 13. (b) Let =λ 9999 − n AC 2 =
⇒
∴
⇒
⇒ ⇒ ⇒ ⇒
9999λ ≤ 16132 + 2019λ 16132 ⇒ λ≤ 7990 8 16132 ≤ λ≤ ∴ 9998 7990 ∴ λ = 1, 2, λ is an integer. For λ = 2, n is not an integer. Hence, only one value is possible. 14. (a) Let side of square ABCD AB = x F
22
D
Q
∴
θ
θ
16. (b) Let m kg of ice is taken. Then, HI = mLf = 334 m J and HII = mc∆T = m (419 . ) (100) = 419 mJ HIII = mLV = m (2260) = 2260 m J ∴ HI < HII < HIII . 17. (c) For additional extension x, work done is 1 1 k (2x)2 − kx2 2 2 1 1 1 W = k (4x2 ) − kx2 = k (3x2 ) 2 2 2 ∴Ratio is 1 : 3 : 5 : .... W =
18. (c) Let x = temperature value. Then,
E
when ⇒
B
1 BQ = BC = CQ 2 1 AP = AB = BP 2
θ A
In ∆PBC, PC 2 = BC 2 + PB 2 1 PC 2 = BC 2 + AB 2 4
∴ m = 2! × 2! × 2! = 8 ways H1 × W1 X n = Number of ways in circular permutation = 2! = 2 ∴ m = 4n
C
θ θ
...(i)
In ∆AQB, 1 ...(ii) AQ 2 = AB 2 + BC 2 4 On adding Eqs. (i) and (ii), we get 5 PC 2 + AQ 2 = (AB 2 + BC 2 ) 4 5AC 2 ⇒ PC 2 + AQ 2 = 4 4 2 ⇒ AC = (PC 2 + AQ 2 ) 5 4 2 ⇒ AC = (192 + 222 ) 5 [Q PC = 19, AQ = 22]
W2 W1
9999λ ≥ 8 + λ 9999λ 8 and ≤ 2019 λ≥ 8+ λ 9998
G
P
H2
n ∈[1, 2019] 9999λ ≤ 2019 1≤ 8+ λ 9999λ ≥1 8+ λ
19
A
H1
8n = 9999λ − λn 9999λ n= 8+ λ
⇒
∴
15. (c) m = Number of ways in which two couples can be seated in 4 chairs in a row such that no wife is next to husband.
B
∴
C− 0 F − 32 = 100 − 0 212 − 32 C = F = x, x − 32 x = ⇒ x = − 40 100 180 − 40°C = − 40°F
In ∆ABE,
19. (a) We have,
BE tan θ = AB BE = x tan θ CE = BC − BE = x(1 − tan θ) In ∆CEF, CE tan θ = ⇒ CF = x(cot θ − 1) CF Now in ∆GHF, FH CE tanθ = = GH DF x(1 − tan θ) CE = = DC − CF x(2 − cot θ) 1 − tan θ tan θ = 2 − cot θ 2 ⇒ 2 tan θ − 1 = 1 − tan θ ⇒ tanθ = 3 2 sinθ = ∴ 13
92 U
238
α →
90 Th
234
−
β →
91 Pa β
234
→ 92 U 234
20. (a) Amplitude of resultant motion is A= =
A12 + A22 + 2A1 A2 cosφ 2+ 2+ 2×
2×
2×
1 = 2
6
So, maximum acceleration = ω2A = 12 ×
21. (d) vr2 = vm2 − v2 v=
320 = 80 m/min 4 vr
vm
v
6=
6
cm s2
202
KVPY Practice Set 3 Stream : SA vm =
5 vr 3
35. (d) HgS, when heated strongly in air gives mercury and sulphur dioxide. θ
2
5 vr2 = vr − (80)2 3 16 2 vr = (80)2 9 80 × 3 vr = = 60 m/min 4 22. (d) X = Mx − mx′ = 4M (0) − M (4) M−m 4M − M 4 =− 3 My − my′ 4M (0) − M (2) 2 =− Y = = 3 M−m 4M − M
Roasting HgS + O2 → Hg + SO2 The process is known as roasting where the sulphide ore is directly heated in presence of air (O2 ) to get the respective metal.
Density of sphere =
−4
4 π × (5) × (10 ) 2
= 44 kg m −3
30. (b) I2 I1
23. (c) For isothermal process, pV = constant ⇒ p = K ⇒ log p = − log V + log K V This is a straight line with negative slope.
24. (c) As emergent beam is parallel, so mirror must forms image at focus of lens.
S
25 cm
Clearly, d = 12 cm. −2 25. (a) [K ] = [ p] [V 5 /3 ] = [MLT2 ] ⋅ [L3 ]5 /3
[L ]
= [ML4 T −2]
26. (c) Electrons which have maximum energies are near to fermi energy level. These electrons can jump to higher energy levels. 27. (b) As smoke rises up in air, speed of smoke increases and flow becomes turbulent. 28. (d) As, F = − dU , option (d) is dr correct.
29. (a) T cosθ = mg and T sinθ = N
At
⇒
26 cm
4 cm
1 1 1 For concave lens using, − = v u f 1 1 1 we have − = ⇒ v = 5 cm v 4 −20 v Magnification of concave lens = = 1.25 u As size of I1 is 2 cm. ∴ Size of I 2 = 2 × 125 . = 2.5 cm
31. (b) Given that, 15 cm
⇒
36. (a) The oxidation state of S in the
. 14
w 13 w = cosθ 12 T = 15 N and w = mg , 13 mg [Q g = 10 ms −2] 15 = 12 15 × 12 = 13 × 10m 12 × 15 . kg ≈ 14 m= 13 × 10 T =
M1 = 5 M, V1 = 500 mL, V 2 = 1500 mL For dilution, M1V1 = M2V 2 5 × 500 = M × 1500 5 M = = 1.66 M 3 32. (c) According to Boyle’s law, at constant temperature, the volume of a given mass of a gas is inversely 1 proportional to its pressure, i.e. p ∝ V As, V1 > V 2 > V3 > V 4 ∴ p1 < p2 < p3 < p4
33. (b) Calcium nitrate gives brick red flame which breaks down on heating to give oxygen and NO2 which is a brown ∆ gas 2Ca(NO3 ) → 2CaO + O2 + 4NO2 34. (a) When but1yne reacts with excess HBr, the major product obtained is 2,2 dibromobutane. This reaction follows Markownikoff’s rule CH3 CH2C≡≡ CH + HBr → Butyne
CH3 CH2 C == CH2 Br
HBr →
Br CH3 CH2 C CH3 Br 2, 2 dibromobutane
given options are as follows (i) Dithionous acid O O HO S S OH 2(x) + 2(−2) + 2(− 1) = 0 2x − 4 − 2 = 0 2x = 6 x=+ 3 (ii) Sulphurous acid
O HO
S
OH
1(x) + 1(− 2) + 2(− 1) = 0 x=+ 4 (iii) Dithionic acid O O HO S S OH O O 2(x) + 2(− 1) + 4(− 2) = 0 2x −2 − 8 = 8 2x = 10 x=+ 5 (iv) Pyrosulphuric acid O O HO S O S OH O O 2(x) + 5(− 2) + 2(− 1) + = 0 2x − 10 − 2 = 0 2x = + 12 x=+ 6 Thus, the correct option is (a).
37. (a) Boiling point of a compound is dependent of on Hbonding present in it. Intermolecular Hbonding is more in primary than in secondary amines as there are two Hatoms available for Hbonding. Tertiary amines do not have intermolecular Hbonding due to the absence of Hatom. Therefore, the order of boiling points of the given amines is as follows nC4 H9 NH2 > (C2H5 )2 NH > C2H5 N(CH3 )2 nbutylamine diethylamine N,N dimethylethylamine (1°) (2°) (3°)
203
KVPY Practice Set 3 Stream : SA 38. (c) Given velocity of light
43. (c) Among the given carbocations
= 3 × 108 ms−1 Frequency of light = 8 × 1015 s−1 C 3 × 108 ms−1 As we know, λ = = ν 8 × 1015 s−1
benzyl and allyl carbocations are more stable than methyl and vinyl carbocations because they have delocalised electrons. An allylic carbocation has two resonance structures whereas in benzylic carbocation has five resonance structures. Thus, benzyl carbocation is the most stable one.
= 3.75 × 10−8 m 1 m = 109 nm −8 ∴3.75 × 10 m = 3.75 × 10−8 × 109 nm = 3.75 × 101 nm ≈ 4 × 101 nm
+
+
RCH==CHCH2
RCH—CH==CH2 (Allylic carbocation)
39. (a) Electron affinities of II period element are less negative as compared to corresponding III period element. This is because of small size of II period elements. Also, nitrogen has the least electron affinity due to stable halffilled configuration. Thus, the correct order of electron affinity is N < O < S < Cl.
+ +
—CHR
==CHR
==CHR
==CHR
+
+
40. (d) Xe atom has 8 electrons in its +
outermost shell. In case of XeF2, out of these 8 electrons, 2 are used for bond formation, while 3 pairs remains nonbonded, i.e. it has 3 lone pairs. In case XeF4 , 4 electrons of Xe are used for bonding. Thus it has 2 lone pairs. In case of XeF6 , 6 electrons are involved for bond formation, thus, it has only 1 lone pair.
==CHR (Benzyl carbocation)
44. (b) 2
CH CH3
CH2
3
5
CH
CH2
Xe
F
Xe
F
F
F
F
Xe F
F XeF2
XeF4
XeF6
(3 lone pairs)
(2 lone pairs)
(1 lone pair)
41. (d) When aqueous solution of +
6
CH3
CH2
Thus, the correct IUPAC name of the given compound is 4ethyl3propyl hex1ene.
Phenol
46. (d) Carbohydrates, fats and proteins
42. (a) The electronic configuration of chromium (Z = 24) is 1s2 2s2 2 p 6 3s2 3 p 6 4s1 3d5 for 19th electron the orbital is 4s Thus, n=4 l = 0 to 3 m = − 3 to 3 1 s=+ 2 Thus, among the given options, only (a) corresponds to the above given values. So, option (a) is correct.
49. (a) Stalled fork activates checkpoint signaling and pauses replication. Since G1 / S checkpoint checks DNA damage, cell size prior to Sphase (i.e. DNA replication phase) this checkpoint would be activated by stalled DNA replication fork.
50. (d) Ester bonds are the least likely to be involved in stabilising the 3D folding of most proteins. A long protein chain gets folded upon itself like a hollow woolen ball, giving rise to a tertiary (3D) structure. This structure is stabilised by several types of bonds, i.e. hydrogen bonds, ionic bonds, van der Waal’s interactions, covalent bonds and hydrophobic bonds. Ester bond is formed between sugar and phosphate in a nucleotide and is not involved in stability of a polypeptide chain. Thus, option (d) is correct. 51. (a) Insulin is the peptide hormone which enhances the uptake of glucose molecules by liver cells (hepatocytes) and fat cells (adipocytes) for its cellular utilisation. Such an activity of insulin brings down the level of glucose in the blood. 52. (a) The cross for the question is XC Y
×
XX
45. (c) Since, each period starts with the
OH
∆ Benzene diazonium chloride
H3C
H2C
filling of electrons in a new principal quantum number, therefore the period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period. Thus, period number = maximum n of any element. (where, n = principal quantum number)
benzene diazonium chloride is boiled, it gives phenol.
N≡≡NCl –
F F F
4
CH
F F
1
CH2
48. (b) The partial pressure of oxygen ( pO2 ) in alveoli of lungs is 104 mm Hg, which is more than that of blood in the blood capillaries of lung alveoli (40 mm Hg). This difference allows passive diffusion of O2 from air filled in the lungs to the blood vessels of lung alveoli.
all can be used as a substrate in cellular respiration. All of them first get converted to acetyl CoA to enter Kreb’s cycle of aerobic cellular respiration. Thus, it is the common factor of respiration entering Kreb’s cycle after breakdown of carbohydrates, fats and proteins.
47. (b) When a tissue with a potential of differentiation is grown in an artificial medium containing auxin and cytokinin in a specific ratio, it starts differentiating. Thus, root and shoot differentiation occurs. Auxin initiates root formation while cytokinin starts shoot formation.
XXC
XCX
XY
Carrier Normal Carrier daughter son daughter
XY Normal son
Since the male offsprings get Xchromosome from their mother who is normal homozygous, thus, none of the son would be colourblind.
53. (d) Cell wall of the most fungi is made up of chitin. Chemically it is Nacetyl glucosamine. It is found in the exoskeleton of insects.
54. (d) Parents TT (Tall)
×
F1generation
tt (Dwarf) Tt (Heterozygous tall On selfing)
T
t
T
TT (Tall)
Tt (Tall)
t
Tt (Tall)
tt (dwarf)
F2generation
Phenotypic ratio 3 : 1 [Tall : Dwarf] Genotypic ratio 1 : 2 : 1
204
KVPY Practice Set 3 Stream : SA
56. (b) Sustained muscle contraction due to repeated stimulus is known as tetanus. This results due to muscle fatigue.
57. (c) Sphincter of Oddi guards the opening of hepatopancreatic duct into the duodenum. Hepatopancreatic duct brings secretion of liver as well as pancreas to the duodenum. 58. (b) Lysosomes are hydrolytic
(x − 1) ( y − 1) = (x, y) + 1 Put x = 2, y = 3 (2 − 1) (3 − 1) = GCD of (2, 3) + 1 2 = 1+ 1 ∴ (2, 3) x = 3, y = 3 (3 − 1) (3 − 1) = 4 = GCD of (3, 3) + 1 x = 2, y = 4 also satisfies When x > 3 not satisfies the equation. ∴ Only 3 pairs (2, 3), (3, 3) and( 2, 4) satisfy the equation. AB and AC respectively. AX 1 AY 2 = and = 1 XB 2 YC Area of ∆AXY 1 = Area of ∆BXY 2
59. (b) Parapodia are present in aquatic
⇒
60. (c) Anthocyanins are water soluble vacuolar pigments that may appear red, purple or blue depending on pH. It is impermeable to cell membranes of plants and can leak out only when membrane is damaged or dead. 61. (a) Let the set S has n elements. ∴Mean of S and {15}
S + 15 m+ 2= n+1
⇒ m + 2n = 13 Also, mean of S and {15, 1} is S + 15 + 1 m + 1= n+ 2 ⇒ (n + 2) (m + 1) = S + 16 = nm + 2m + n + 2 = nm + 16 ⇒ n + 2m = 14 From Eqs. (i) and (ii), we get n=4 ∴S has 4 elements.
A
⇒
I Q
Area of ∆ABY = Area of ∆AXY + Area of ∆BXY = 10 + 20 = 30 A 2
X
Y 2
1
B
C
⇒
Area of ∆ABY 2 = Area of ∆CBY 1
∴
Area of ∆CBY =
...(i)
30 = 15 2
66. (c) By homogenity principle,
= 30 + 15 = 45
⇒
64. (a) We have, x1 + x2 + x3 + K + x2019 = 1 x x2 x2019 and 1 + + K+ =1 1 − x1 1 − x2 1 − x2019
...(ii)
2 x12 x22 x32 x2019 + + + K+ 1 − x1 1 − x2 1 − x3 1 − x2019
x2 − x1 + x1 x2 − x2 + x2 = 1 + 2 +K 1 − x1 1 − x2 2 x2019 − x2019 + x2019 1 − x2019 x1 (x1 − 1) x1 x2 (x2 − 1) = + + + x2 1 − x1 1 − x1 1 − x2
+
+ K+
x2019 x2019 − 1) 1 − x2019
C
In ∆IBC, ∠BIC + ∠IBC + ∠ICB = 180° ∠B ∠C ∠ BIC + + = 180° 2 2 ∠B + ∠C ∠BIC = 180° − 2 180 − ∠ A = 180° − 2 ∠A = 90° + = 90 + 20 = 110 2 [Q ∠A = 40° ] ∠BIC = ∠QIR = 40° 1 ∠QPR = QIR 2 [Q angle in a segment is half of angle on a centre segment of circle] 1 ∠QPR = × 110° = 55° 2
+ Area of ∆CBY
⇒
R
AI, BI and CI intersect the circle at P, Q, R respectively.
Area of ∆BXY = 20
1
P
B
[Q area of ∆AXY = 10]
Q m = S n
62. (b) We have, (x, y) = GCD of x and y. Given, xy = x + y + (x, y) xy − x − y = (x, y) xy − x − y + 1 = (x, y) + 1
In ∆ABC, I is incentre of ∆ABC.
∴ Area of ∆ABC = Area of ∆ABY
⇒ (n + 1) (m + 2) = S + 15 (n + 1) (m + 2) = nm + 15
= −1+ 1= 0
65. (c) Given,
63. (b) In ∆ABC, X and Y are points on
enzymes containing cell organelles which are bounded by a single membrane. Other organelles like chloroplast, mitochondria and nuclei have double membrane system. animals, i.e. annelids like Nereis which help them in swimming. Other three features, i.e. metameric segmentation, jointed appendages and chitinous exoskeleton are present in phylum Arthropoda. Out of these, metameric segmentation is visible as tagmetisation.
= − (x1 + x2 + K + x2019 ) x x2 x2019 + 1 + + K+ 1 1 1 − − − x x x2019 1 2
°
increase in acidity favours the dissociation of oxyhaemoglobin thereby giving up more O2. When this phenomenon occurs due to increase in CO2 concentration, then it is called Bohr effect.
40
55. (b) Reduction of pH of blood, i.e.
+
x2019 1 − x2019
[a ] = dimensions of power W = [ML2 T −3 ] T [bt 2 ] = dimensions of power [ML2T −3 ] [b] = = [ML2 T −5 ] [T 2 ]
[c] = dimensions of t3 = [T3 ] [d t ] = dimensions of power [T3 ] ⇒ = [M−1 L−2 T −6 ] [d ] = [ML2T −3 ] −1 3
So, statement IV is incorrect.
67. (b) Before collision, velocity of translation of sphere = vcube = rω = 1 cms−1 = 0.01 ms−1 As collision is elastic, translational kinetic energy of sphere is transferred to the cube but its rotational kinetic energy remains constant. ∴ After collision, vsphere = 0, vcube = 0.01 ms−1 and ωsphere = 1rad s−1 .
205
KVPY Practice Set 3 Stream : SA 71. (d)
68. (b) By KVL,
C12H22O11 (s) + 12O2 ( g ) → 12CO2 ( g )
a 2kΩ A
1kΩ 1–a
b D a+b C 1kΩ 1kΩ
+ 11H2O(l) ∆H C° = [12∆ f H ° (CO2 ) + 11∆ f H ° (H2O)] − [∆ f H ° (C12H22O11 )]
B
1–a–b 2kΩ
we have, V AD = 2a = 1 − a + b Also, VCD = 2 (1 − a − b) = b + a + b 2 a= ∴ 5 1 and b= 5 Hence, current through 2 kΩ resistor 2 = a = mA. 5 and current through middle 1 kΩ resistor 1 = b = mA. 5
= [12(−94.3) + 11 (− 68.3)] − [− 530] = − 1352 .9 J kcal mol −1 Thus, number of moles of C12H22O11 required for 2700 kcal of energy 2700 = ≈ 2 mol = 682.6 g 1352.9
72. (d) The alcohol on treatment with acid gives an alkene (A) which on ozonolysis will give nonan2, 8 dione. The reaction can be shown as CHOHMe
YyRr
3
=
−2 x2 + 4
+
4x
(x2 + 4)2
2(x2 + 4) = 4x2
⇒ 2x2 = 8 ⇒ x=± 2 As U is minimum at x = + 2 . − dU =0 F= ∴ dx Now, restoring force constant. k=
∴
d 2U dx
2
= at x = 2
ω=
k = m
=
100 4
1 units 8
1 = 8m
= 5 rad s−1
2
1 −3 8 × 5 × 10
YyRr Gamets YR yR Yr yr
YR
yR
Yr
yr
YR
YYRR
YyRR
YYRr
YyRr
yR
YyRR
yyRR
YyRr
yyRr
Yr
YYRr
YyRr
YYrr
Yyrr
yr
YyRr
yyRr
Yyrr
yyrr
1
O
5
dU 2 4x2 = 0 when 2 = 2 dx x + 4 (x + 4)2 ⇒
YR yR Yr yr
O3
4
2
×
A
2mv 2mv Force on Utube = = πd ∆t 2v
≈ 0.23 N = 0.2 N 70. (a) As, U = 2−2x x + 4 (− 2x) (− 1) (2x) −2 dU + = 2 ⇒ dx (x2 + 4)2 x + 4
76. (b) The nerves are myelinated with unmyelinated segments called nodes of Ranvier. The high phospholipid content of the myelin sheath offers electrical insulation, thus saltatory conduction occurs as impulse jumps from one node to the next. This form of conduction facilitates a very rapid transmission of impulses.
Me
ball = − 2mv
4 mv2 4 × 0.02 × 9 = πd 314 . ×1
equation ∆G° = − 2.303 RT log K sp ∆G° = + 63.3 kJ = 63.3 × 103 J 63.3 × 103 = − 2.303 × 8.314 × 298 × log Ksp log K sp = − 11.09 K sp = 8.0 × 10−12
77. (a) The cross for the question is
+ H
69. (b) Change in momentum of steel
=
75. (b) ∆G° is related to K sp by the
8 9 6 7 Nonan2, 8 dione
73. (c) Number of equivalents of silver formed = number of equivalents of copper formed. In AgNO3 , Ag is in + 1oxidation state. In CuSO4 , Cu is in + 2 oxidation state. 108 = 108 ∴ Equivalent weight of Ag = 1 Equivalent weight of Cu 63.6 = = 31.8 2 Weight of silver Eq. wt of silver = Weight of copper Eq. wt of copper 10.79 108 = wCu 31.8 10.79 × 31.8 wCu = 108 = 3.2 g hc 74. (c) Energy of one photon = λ 6.626 × 10−34 Js × 3 × 108 ms−1 = 550 × 10−9 m = 3.61 × 10−19 J ∴ Number of photons energy required = energy of one photon =
10−17 3.61 × 10−19
= 27.67 ≈ 28
The given Punnett square shows 9 : 3 : 3 : 1 ratio of the phenotypes only.
78. (b) As the substrate concentration increases, the rate of reaction increases until a maximum, when saturation of all the enzymes active sites occurs. When this happens, the limiting factor is enzyme concentration. Thus, graph (b) is correct. 79. (c) Average amino acid residues molecular weight = 110 Polypeptide chain of molecular weight 20,000 20,000 = = 182 amino acids 110 A triplet of bases in the DNA molecule codes for one amino acid in a polypeptide chain. To translate 182 amino acids, there must be a minimum of 182 × 3 = 546 nucleotides. 80. (b) Sequoia is one of the tallest tree species, known as red wood tree. It is a gymnospermic plant. Salvinia is an angiosperm, but Ginkgo and Pinus are gymnosperms. Gymnosperms are welladapted to extremes of climate and are heterosporous.
206
KVPY Practice Set 4 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
PRACTICE SET 4 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) MATHEMATICS
5. If a , b, c ≥ 4 are integers, not all equal and
1. A natural number K is such that K 2 < 2019 < (K + 1)2. Then, the largest prime factor of K is (a) 11
(b) 13
(c) 7
(d) 5
2. If real number a, b, c, d, e satisfy a +1=b+2=c+3=d +4=e+5 = a + b + c + d + e + 3, then the value of a 2 + b2 + c2 + d 2 + e2 is equal to (a) 8
(b) 9
(c) 10
(d) 11
3. Let a semicircle with centre O and diameter AB. Let P and Q be points on the semicircle and R be a point on AB extended such that OA = QR < PR if ∠POA = 102°, then ∠PRA is equal to (a) 51° (c) 25.5°
(b) 34° (d) None of these
4. If x = cos 1° cos 2° cos 3° K cos 89° and y = cos 2° cos 6° cos 10° K cos 86°, then the integer 2 y nearest to log 2 is x 7 (a) 16
(b) 17
(c) 18
(d) 19
4abc = (a + 3) (b + 3) (c + 3), then (a + b + c) is equal to (a) 14
(b) 15
(c) 16
(d) 18
6. In a ∆ABC, right angled at A, the altitude through A
and the internal bisector of ∠A have lengths 3 and 4 respectively. Then, the length of median through A is
(a) 20
(b) 24
(c) 15
(d) 10
7. A rectangular floor that is 10 feet wide and 17 feet long is tiled with 170 onefoot square tiles. A bug walks from one corner to the opposite corner in a straight line including the first and the last tile, how many tiles does the bug visit? (a) 17
(b) 25
(c) 26
(d) 27
8. Ashwani computes the mean µ, the median M and the modes of the 365 values that are the dates of 2019. Thus his data consists of 12 1s, 12 2s ..., 12 28s, 11 29s, 11 30s and 7 31s. Let d be the median of modes. Which of the following is correct? (a) µ < d < M (c) M = d = µ
(b) M < d < µ (d) d < µ < M
207
KVPY Practice Set 4 Stream : SA 9. A sequence of numbers is defined recursively by
v (ms–1)
an − 2 ⋅ an − 1 3 for all n ≥ 3. a1 = 1, a 2 = and a n = 2a n − 2 − a n − 1 7 p Then, a 2019 can be written as , where p and q are q relatively prime number, then the value of p + q is equal to
(a) 6057
(b) 8087
(c) 8078
15
3
Now, consider the following statements:
(d) 4039
I. Force acting on particle is 50 N.
10. A child builds towers using identically shaped cube of
II. Force stops at t = 3 s.
different color. Then, number of different tower with a height 8 cubes can the child build with 2 red cubes, 3 blue cubes and 4 green cubes (one cube is left out) is (a) 24
(b) 288
(c) 312
III. Force stops at t = 4 s. IV. Particle receives an impulse at t = 4 s. Which of the above statements are correct? (a) Statements II and IV are correct (b) Statements I, II and IV are correct (c) Statements I, III and IV are correct (d) Statements III and IV are correct
(d) 1260
11. The least possible value of
(x + 1) (x + 2) (x + 3) (x + 4) + 2019 is (where, x is real)
(a) 2017
(b) 2018
(c) 2019
(d) 2020
12. Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points A and B, as shown in the figure. The m distance AB can be written in the form , when m n and n are relatively prime.
17. Correct graph of experimental values of specific heat of a constant volume of hydrogen gas is (b)
(a) CV 3 —R 2
(c)
3/2 R
(b) 29
(c) 69
(d) 58
completing one round in 40 s. Another person Y running in opposite direction meets X every 15 s. The time, expressed in seconds, taken to Y to complete one round is (b) 24
T(K)
600
CV 7/2 R
(c) 25
(d) 55
3/2 R
3/2 R
13. A person X is running around a circular track
(a) 12.5
(d)
CV
80
5/2 R
Then, m + n is (a) 21
3000
T(K)
7/2 R
B
CV
5/2 R
80
A
t(s)
4
80
T(K)
3000
80
T(K)
600 3000
139 18. Ratio of nuclear density of nuclei 142 53 I and 56 Ba is
(a) 142 : 139 (c) 139 : 142
(b) 53 : 56 (d) None of these
19. A particle starts from origin, it accelerates first
where b is odd. Then, number of all such 6digit numbers that are divisible by 7 is
t0 second and then deaccelerates at same rate till 2 t0 second along the positive xdirection. Variation of displacement x with time t for the particle is given by
(a) 70
(a) x
14. Consider all 6digit numbers of the form abccba, (b) 80
(c) 75
(d) 85
(b) x
15. Let ABCD be trapezium in which AB is parallel to CD and AD is perpendicular to AB. Suppose ABCD has incircle which touches AB at Q and CD at P. Given that PC = 36 and QB = 49, then length PQ is (a) 85
(b) 84
(c) 76
t0
(d) 80
2t0
t (d)
(c) x
t0
2t0
t
x
PHYSICS 16. Velocitytime graph of a particle of mass 10 kg pushed along a frictionless surface by an external force is as shown below.
t0
2t0
t
t0
2t0
t
208
KVPY Practice Set 4 Stream : SA
20. A mass m initially at rest is pulled with a force F.
15Ω A
If force is proportional to instantaneous time t, then kinetic energy of the particle is proportional to (b) t −2
(a) t 2
(c) t 4
21. Density of ice is x (g/cc) and that of water is g (g/cc). Change in volume in cc when m grams of ice completely melts is (a) m ( y − x)
(b) ( y − x) / m 1 1 (d) m − y x
(c) mxy ( y − x)
22. How much work is required in units of electronvolt to carry an electron from the positive terminal of a 12 V battery to the negative terminal in external circuit? (b) − 12 eV (d) − 6 eV
(a) 12 eV (c) 6 eV
R1
(d) t 0
2Ω
3Ω
B
(a) 20Ω (c) 5Ω
(b) 10Ω (d) 25Ω
26. Fermi energy level for an electron is (a) a possible energy value that an electron can have in free state (b) an unfilled energy level that can be occupied by two electrons of opposite spins (c) lowest energy value possible for a bound electron (d) highest occupied energy level at absolute zero kelvin upto which every possible energy levels are filled
27. Solar cookers are not very popular because
23. Following graph shows atmospheric pressure, gauge pressure and absolute pressure. p
(a) they are bulky (b) they are not put into kitchen (c) they cook food in large time (d) sun changes its position rapidly
28. A cubical block of side 1 m and mass 10 kg is placed on a rough surface. Block can be toppled by applying a force horizontally at its upper edge. Minimum value of F is
p2 p3 p1
F
Then, choose the correct option. (a) p1 p3 (b) p1 p3 (c) p1 p3 (d) p1 p3
= gauge pressure, p2 = atmospheric pressure, = absolute pressure = atmospheric pressure, p2 = gauge pressure, = absolute pressure = absolute pressure, p2 = atmospheric pressure, = gauge pressure = gauge pressure, p2 = absolute pressure, = atmospheric pressure
(a) 100 N (c) 50 N
(b) 200 N (d) 25 N
29. Potential energy between two molecules as a function of their separation is as shown below. U
24. A man can walks on hard ground with a speed of
5 ms−1 and on sandy ground with 3 ms−1. Let he is standing on border of sandy and hard ground and wishes to reach the tree situated on the sandy ground as shown below. D Man
O
0.6 Å 0.4 Å
x
1.2 Å
Force between particles is zero at (a) x = 0.4 Å (c) x = 12 . Å
Tree
1.8 Å
(b) x = 0.6 Å (d) x = 18 . Å
30. Using following figures,
120 m E
A
100 m
B C
He can reach tree in least time when he walks on hard ground upto point E and then he walks straight towards tree along EO. Distance AE is (a) 10 m (c) 30 m
(b) 20 m (d) 50 m
25. Equivalent resistance between A and B is 6 Ω. Value of resistance R1 is
µ1
µ
µ1
µ2
µ
µ2
Relation between refractive indices µ1 and µ 2 is
(a) µ 1 < µ 2 (c) µ 1 > µ 2
(b) µ 1 µ 2 = µ 2 (d) µ 1 = µ 2
209
KVPY Practice Set 4 Stream : SA Cu 2+ / Cu (s), E° = + 034 . . I2 (s) / I− , E° = + 054
CHEMISTRY 31. How many moles of magnesium phosphate
(a) Cu will reduce Br − (c) Cu will reduce I−
Mg 3 (PO 4 )2 will contain 0.25 mole of oxygen atom? (b) 3125 . × 10−2 (d) 2.5 × 10−2
(a) 0.02 (c) 125 . × 10−2
39. The carboxylic acid which reduces Tollen’s reagent is
32. At what temperature will the r.m.s velocity of SO 2 be the same as that of O 2 at 303 K ?
(a) 350 K (c) 606 K
is treated with acetic acid and lead acetate, a black precipitate is obtained. This suggests that the organic compound contains (b) phosphorus (d) nitrogen
presence of hydrochloric acid. The product obtained is
O (a) CH3 CH2CH2 C CH3 O (b) CH3 CH2CH2 C CH2CH2CH3
(c) 24 g
(d) 0 g
15 . h π
(b)
6h π
(c)
3h π
(d)
3h π
and then with NaF. When gaseous BF3 is passed through the solution obtained, a precipitate X is formed. The formed pricipitate X is
OC2H5
(a) Na3[AlF6 ] (b) Na[BF4 ]
OC2H5
(d) H3 [AlF6 ]
(c) AlF3
45. Which one of the following will be aromatic? +
(a)
(b)
(b) N < Si < C < P (d) P < Si < N < C
kg moving with a velocity of 10 ms
−1
is
(b) 6.63 × 10−16 m (d) 6.63 × 10−29 m
37. The product formed when 1bromo3chlorocyclobutane reacts with two equivalents of metallic sodium in ether, is Cl (b) Br (c)
(b) 16 g
44. Aluminium trifluoride is treated with anhydrous HF
36. The deBroglie wavelength associated with particle of
(a)
mass of O 2 must be released to reduce the pressure in the flask to 12.315 atm?
(a)
increases in the order
mass of 10
(b) 2 moles of HCl ( g ) (d) 1.5 moles of HCl ( g )
in forbital ?
OC2H5
(a) 6.63 × 10−7 m (c) 6.63 × 10−21 m
each at STP, the moles of HCl( g) formed is equal to (a) 1 mole of HCl ( g ) (c) 0.5 mole of HCl ( g )
43. What is the orbital angular momentum of an electron
35. The electronegativity of the following elements
−6
(b) CO < CO32− < CO2 (d) CO < CO2 < CO32−
41. When 22.4 L of H2 ( g) is mixed with 11.2 L of Cl 2 ( g)
(a) 8 g
OH
(a) C < N < Si < P (c) Si < P < C < N
(a) CO2 < CO32− < CO (c) CO32− < CO2 < CO
42. A 1 L flask contains 32g of O 2 gas at 27°C. What
34. Acetone is treated with excess of ethanol in the
(d) (CH ) C 3 2
(b) oxalic acid (d) lactic acid
CO, CO32− , CO 2 is
33. When acidified sodium extract of organic compound
(c) (CH3)2C
(a) acetic acid (c) formic acid
40. The correct order of C—O bond length among
(b) 505 K (d) 100 K
(a) chlorine (c) sulphur
(b) Cu will reduce Ag (d) Cu will reduce Br2
(d)
38. E° values of some redox couples are given below. On the basis of these values choose the correct option. . Br2 / Br− , E ° = + 190 + Ag / Ag (s), E° = + 080 .
(c)
(d)
ρ
σ
BIOLOGY 46. Which cells of ‘crypts of Lieberkuhn’ secrete antibacterial lysozyme? (a) Argentaffin cells (c) Zymogen cells
(b) Paneth cells (d) Kupffer cells
47. Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen? (a) Bacillus (c) Mycoplasma
(b) Pseudomonas (d) Nostoc
48. The cell organelle responsible for extracting energy from carbohydrates to form ATP is (a) lysosome (c) chloroplast
(b) ribosome (d) mitochondrion
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KVPY Practice Set 4 Stream : SA
49. DNA fragments are
55. Fruit and leaf drop at early stages can be prevented
(a) positively charged (b) negatively charged (c) neutral (d) either positively or negatively charged depending on their size
50. An important characteristic that hemichordates share with chordates is
by the application of (a) cytokinins (b) ethylene (c) auxins (d) gibberellic acid
56. Which one of the following options best represents enzyme composition of pancreatic juice?
(a) absence of notochord (b) ventral tubular nerve cord (c) pharynx with gill slits (d) pharynx without gill slits
(a) Amylase, peptidase, trypsinogen, rennin (b) Amylase, pepsin, trypsinogen, maltase (c) Peptidase, amylase, pepsin, rennin (d) Lipase, amylase, trypsinogen, procarboxypeptidase
51. Lungs are made up of airfilled sacs, the alveoli. They do not collapse even after forceful expiration, because of (a) Residual Volume (RV) (b) Inspiratory Reserve Volume (IRV) (c) Tidal Volume (TV) (d) Expiratory Reserve Volume (ERV)
57. In the fruit fly, Drosophila melanogaster, the diploid number of chromosomes is 8. In the absence of crossing over or mutation, how many genetically unique kinds of gamete might be formed by one individual? (a) 4 (c) 16
52. Viroids differ from viruses in having
(b) 8 (d) 32
58. Biochemical analysis of a sample of DNA shows that cytosine forms 40% of the nitrogenous bases. Which percentage of the bases is adenine?
(a) DNA molecules with protein coat (b) DNA molecules without protein coat (c) RNA molecules with protein coat (d) RNA molecules without protein coat
(a) 10% (c) 40%
53. Plants, which produce characteristic pneumatophores
(b) 20% (d) 60%
59. The first stable product of fixation of atmospheric
and show vivipary belong to
nitrogen in leguminous plant is
(a) mesophytes (c) psammophytes
(a) NO−2 (c) NO3−
(b) halophytes (d) hydrophytes
60. Treponema pallidum pathogen is a cause of
54. Spliceosomes are not found in cells of (a) plants (c) animals
(b) ammonia (d) glutamate
(a) leprosy (c) syphilis
(b) fungi (d) bacteria
(b) plague (d) pertussis
PARTII (2 Marks Questions) MATHEMATICS 61. Integer a, b, c satisfy a + b − c = 1 and a 2 + b2 − c2 + 1 = 0, then the sum of all possible values of a 2 + b2 + c2 is equal to (a) 17
(b) 18
(c) 20
(d) 24
62. There are several tea cups in the kitchen, some with handles and others without handles. The number of ways of selecting two cups without a handle and three with a handle is exactly 1200. Then, the maximum possible numbers of cups in the kitchen is equal to (a) 25
(b) 27
(c) 28
(d) 29
63. Let D be an interior point of the side BC of a ∆ABC. Let I1 and I 2 be the incenters of ∆ABD and ∆ACD
respectively. Let AI1 and AI 2 meet BC in E and F respectively. If ∠BI1E = 60°, then ∠ CI 2F is (in degree) (a) 30°
(b) 45°
(c) 75° 2
(d) 60°
64. Let P (x) = a 0 + a1x + a 2x + K + a n x be a polynomial n
in which a i is a nonnegative integer for each i ∈(0, 1, 2, 3, K , n ). If P(1) = 4 and P(5) = 136, then P(3) is (a) 25
(b) 30
(c) 32
(d) 34
65. In a quadrilateral ABCD, it is given that
AB = AD = 13, BC = CD = 20, BD = 24. If r is the radius of the circle inscribed in the quadrilateral, then the integer closest to r is
(a) 6
(b) 8
(c) 9
(d) 10
211
KVPY Practice Set 4 Stream : SA 69. Velocitytime graph of an object moving along a
PHYSICS
straight line is as shown below.
66. Two earthworms climb over a rough thin wall of an
v (ms–1)
earthen pot 10 cm high placed in a lawn.
8
Earthworm
9
Wall
3
One of the worm is 20 cm long and other is only 10 cm long and mass of both earthworms is 20 g. Ratio of work done by worms when they crosses half of their length across top of the wall is (a) 1 : 1 (c) 2 : 1
(b) 2 : 3 (d) 1 : 2
67. A jar of height 20 cm is filled with water (nw = 4 / 3). At centre of jar on the bottom surface, a red dot is made.
t(s)
7
–8
If x = displacement (in m) and a = acceleration (in ms−2). Then, correct graph is (b) x(m)
(a) x(m) 28
a (c) (ms–2)
t(s)
9
t(s)
9
Disc
5
a (d) (ms–2)
2.67
Water Dot
9
(b) 23 cm (d) 2 cm
68. Consider arrangements A and B for making a torch:
9
3
Minimum radius of an opaque plastic disc that makes the dot invisible from top is (a) 20 cm (c) 12 cm
5
t(s)
t(s)
70. A vessel has a hole of radius r = 1 cm. Vessel is initially full of water and hole is sealed by a ball of mass m = π g. Depth of water is now slowly reduced using a syphon and when it reaches a certain value h0, the ball rises out of the hole.
+ – + –
–
+
–
+ (B)
(A)
Now, consider the following statements. I. Torch A is brighter.
Value of h0 is (Radius of ball is slightly larger than hole but for calculation both can be taken same, g = 10 ms−2) (a) 65 cm
(b) 72 cm
(c) 84 cm
(d) 110 cm
II. Torch B is brighter. III. Torch A lasts longer. IV. Torch B lasts longer. Which of the above statements are correct? (a) Statements I and III are correct (b) Statements II and IV are correct (c) Statements I and IV are correct (d) Statements II and III are correct
CHEMISTRY 71. A solid compound X on heating gives CO 2 gas and a residue. The residue mixed with water form Y. On passing an excess of CO 2 through Y in water, a clear solution Z is formed. On boiling Z, compound X is reformed. The compound X is (a) Ca(HCO3 )2
(b) CaCO3 (c) Na2CO3
(d) K 2CO3
212
KVPY Practice Set 4 Stream : SA
72. Standard entropies of X 2 , Y 2 and XY3 are 60, 40 and −1
77. Haemophilia is caused by a sexlinked, recessive allele. A couple have a haemophilic son, a normal son and a haemophilic daughter. What are the most likely genotypes of the parents?
−1
50 JK mol , respectively. For the reaction, 1 3 X + Y XY3 ; ∆H = − 30 kJ 2 2 2 2 To be at equilibrium, the temperature should be

(a) 750 K
(b) 1000 K
(c) 1250 K
H 2 /Ni
Mother Father (a) XH X
(d) 500 K
(b) XH Y
O / H O/ Zn
3 2 73. A (C 4 H6 ) → B (C 4 H8 ) → CH3CHO
H
(c) X X
1 mol
relation to the endocrine system?
74. In the Kjeldahl’s method for the estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralised 10 mL of 1 M H2SO 4. The percentage of nitrogen in the soil is (c) 35.33
(d) 45.33
75. A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. The molecular formula of the compound is
79. If the nucleus of a human motor neuron contains 6.8 picograms (pg) of DNA, what mass of DNA is the nucleus of an actively dividing human skin cell likely to contain at the end of interphase? (a) 3.4 pg (c) 13.6 pg
[Atomic weight of H = 1008 . , C = 12.0 and Br = 799 .] (c) C2 H 4 Br2
(b) CH 2 Br2
(a) CH3 Br
XH Y
(a) Adenohypophysis is under direct neural regulation of the hypothalamus (b) Organs in the body like gastrointestinal tract, heart, kidney and liver do not produce any hormones (c) Nonnutrient chemicals produced by the body in trace amount that act as intercellular messenger are known as hormones (d) Releasing and inhibitory hormones are produced by the pituitary gland
(c) CH 3 CH 2 C ≡≡ CH and CH 3 CH == CHCH 3 (d) CH 2 ==CHCH == CH 2 and CH3 CH == CHCH3
(b) 45.33
XH Y
H
78. Which of the following statements is correct in
and
(a) 37.33
XH Y H
(d) X Y H
Identify A and B in the above reaction. and (a) (b)
XH Y
(d) C2 H 3 Br3
(b) 6.8 pg (d) 20.4 pg
80. The diagram represents a reaction with and without an enzyme. What is the activation energy of the enzyme catalysed reaction?
BIOLOGY 76. The complete oxidation of one mole of glucose yields
(a) 23%
(b) 36%
(c) 40%
Energy
2880 kJ of energy. The addition of one phosphate molecule to ADP requires 30.6 kJ of energy per mole. In aerobic respiration, 38 molecules of ATP are formed as a result of the breakdown of each glucose molecule. Which figure best represents the efficiency of aerobic respiration in trapping the energy released by the glucose molecule?
A
D B
Reactants
Product C Reaction
(a) A (c) C
(d) 45%
(b) B (d) D
Answers PARTI 1 11 21 31 41 51
(a) (b) (d) (b) (a) (a)
2 12 22 32 42 52
(c)
62 72
(d)
(c) (a) (c) (b) (d)
3 13 23 33 43 53
(d)
63 73
(a)
(b) (b) (c) (d) (b)
4 14 24 34 44 54
(d)
64 74
(d)
(a) (a) (d) (b) (d)
5 15 25 35 45 55
(c)
65 75
(b)
(b) (b) (c) (d) (c)
6 16 26 36 46 56
(b)
66 76
(b)
(b) (d) (d) (b) (d)
7 17 27 37 47 57
(c)
67 77
(b)
(d) (d) (d) (c) (c)
8 18 28 38 48 58
(d)
68 78
(c)
(d) (c) (d) (d) (a)
9 19 29 39 49 59
(c)
69 79
(d)
(c) (a) (c) (b) (b)
10 20 30 40 50 60
(d)
70 80
(a)
(c) (a) (d) (c) (a)
PARTII 61 71
(b) (b)
(a)
(d)
(a)
(c)
(c)
(a)
(c)
(c)
(b)
213
KVPY Practice Set 4 Stream : SA
Solutions 1. (a) We have,
=
K 2 < 2019 < (K + 1)2 2
∴
K < 2019
⇒
K
136 ∴a4 = a5 = a 6 K = an = 0 ...(i) ∴ a0 + 5a1 + 25a2 + 125a3 = 136 ∴a3 can be 0 or 1 only Now, P (1) = a0 + a1 + a2 + a3 = 4 ...(ii) If a3 = 0, then a0 + 5a1 + 25a2 ≤ 4 + 20 + 100 = 124 < 136 If a3 = 1 ⇒ a0 + 5a1 + 25a2 = 11 [from Eq. (i)] ⇒ a2 = 0 ⇒ a0 + 5a1 = 11 Also from Eq. (ii), a0 + a1 = 3 ⇒ a1 = 2, a0 = 1 Hence, P (x) = 1 + 2x + x3 ∴ P (3) = 1 + 6 + 27 = 34
2.5 cm 7.5 cm
5 cm
For 10 cm worm centre of mass is raised upto height of 7.5 cm, while for 20 cm worm height of centre of mass is 5 cm from ground. So, ratio of work done by 20 cm worm to that of 10 cm worm is W1 mgh1 5 50 = = = = 2: 3 W2 mgh2 7.5 75
67. (b) Let d = diameter of disc. Spot is invisible, if incident rays from dot d reaching top surface at are at the 2 critical angle. 1 Then by sin ic = µ d /2 and = tan i h 2 × 20 2h d= ⇒ = ≈ 46 cm 2 16 µ −1 −1 9 So, minimum radius of disc = 23 cm.
68. (c) In case A, voltage across bulb is higher, so lamb will burn brighter. In case B, voltage is same as either of battery but each battery supplies only half of current, hence the batteries will lasts twice as long. 69. (d) From vt graph, For 0 < t < 3 s, 8 ms−1 a= = 2.67 ms−2 3s
=
32(32 − 20) (32 − 20) (32 − 24)
For 3s < t < 5s, a = 0 For 5s < t < 9s, −16 ms−1 a= = − 4 ms−2 4s
=
32 × 12 × 12 × 8
70. (a) Forces on the ball are
= 192 ∴Area of quadrilateral ABCD = Area of ∆ABD + Area of ∆BCD = 60 + 192 = 252 Radius of incircle Area of quadrilateral = Semiperimeter of quadrilateral 252 = = 7.63 33 ∴ Nearest integer of r is 8.
5 cm
(i) weight of ball = mg (ii) weight of fluid column above ball = πr 2ρgh 2 (iii) Buoyant force = πr3 ρg 3 2 3 when mg = πr ρg − πr 2ρgh, ball will 3 tend to rise corresponding height h0 of water in vessel is given by 2 mg = πr 2ρg − πr 2ρgh0 3
218
KVPY Practice Set 4 Stream : SA
Substituting given values, we get 2 . 2 01 π × 10−3 × 10 = × π × × 1000 × 10 100 3
Thus, the correct option is (d). H 2 / Ni → CH2 == CH CH == CH2 1, 4addition
( A)
CH3 CH == CH CH3 (B)
2
1 − π × 1000 × 10 × h0 100
O3/H2O/Zn
1 2 = − h0 ⇒ 100 3 ⇒
2CH3CHO
h0 = (0.66 − 0.01) m = 65 cm
71. (b) Compound X is CaCO3 ∆ → CaO + CO2 ↑ CaCO3 Residue
X
CaO + H2O → Ca(OH)2
Residue
Y
Ca(OH)2 + CO2 + H2O → Ca(HCO3 )2 Y
Excess
∆ Ca(HCO3 )2 → CaCO3 + H2O + CO2 ↑ X
72. (a) For the reaction, 1 3 X 2 + Y2 2 2

XY3 (∆H = − 30 kJ)
1 3 ∆S° = ∆S(°XY3 ) − SX° 2 + SY°2 2 2 3 1 = 50 − × 60 + × 40 2 2 = 50 − [30 + 60] = 50 − 90 = − 40 JK−1 mol −1 Also, ∆G ° = ∆H ° − T∆S° At equilibrium, ∆G° = 0 ∆ H = T∆ S ∆H ° T = ∆S ° − 30 × 103 J mol −1 = 750 K = − 40 × JK−1 mol −1 H / Ni
2 73. (d) (A) (C4 H6 ) → (C4 H8 )
1 mole O3/ H 2O/ Zn
→ CH3 CHO By considering the molecular formula C4 H6 . We can conclude that, it is an alkene, CH2 == CH CH == CH2 which on reduction with H2 gives CH3 CH == CH CH3 . Also ozonolysis ofCH3 CH == CHCH3 will only give 2 moles of acetaldehyde.
74. (a)In Kjeldahl’s method, percentage of N is given by 1.4 × normality of acid × volume of acid = weight of compound Also, 1 M H2SO4 = 2 N H2SO4 [M = N × Basicity/Acidity] 1.4 × 2 × 10 ∴% of N = = 37.33 % 0.75
75. (c) Element % of Atomic element weight
No. of moles
Simple ratio
C
12.8
12
12.8/12 1.06 / = 1.06 1.06 = 1
H
2.10
1
210 . / 1 2.10 / = 2.10 1.06 =2
Br
85.1
80
851 . / 80 1.06 / = 1.06 1.06 = 1
Hence, the empirical formula becomes CH2Br Empirical weight of CH2Br = 12 + 2 + 80 = 94 As we know that, molecular wt. n= empirical wt. 187.9 = =2 94 Thus molecular formula = n × empirical formula = 2 × (CH2Br) = C2H4 Br2
76. (c) The total amount of energy used in forming 38 ATP is 38 × 30.6 kJ = 1162.8 kJ Thus efficiency of aerobic respiration is 1162.8 × 100% = 40% 2800 77. (a) Since there is both a haemophilic and normal son, the mother must have a heterozygous genotype. Since there is a haemophilic daughter, the Xchromosome from the father must have the recessive allele. XhX
XhY
XhXh XhY XhX XY Carrier Normal Haemophilic daughter daughter son Haemophilic son
78. (c) Endocrine cells are present in different parts of the gastrointestinal tract, e.g., gastrin, secretin, GIP. Atrial wall of our heart secretes a peptide hormone called ANF (Atrial Natriuretic Factor). Releasing and inhibitory hormones are released by hypothalamus. Adenohypophysis is not under direct control of hypothalamus.
79. (c) The amount of DNA in the 2 somatic cells, the motor neuron and the skin cell should be the same. After interphase where DNA has already replicated, the amount of DNA is doubled, i.e. 6.8 × 2 pg = 13.6 pg.
80. (b) The reaction that is enzymecatalysed has lower activation energy and is represented by the dotted line. The energy input required to raise the energy of the reactants to a certain level before the reaction is triggered is called the activation energy. This is represented by the increase in energy of the reactants to the top of the ‘hill’, B.
219
KVPY Practice Set 5 Stream : SA
KVPY
KISHORE VAIGYANIK PROTSAHAN YOJANA
PRACTICE SET 5 Stream : SA MM 100
Instructions There are 80 questions in this paper. This question paper contains two parts; Part I and Part II. There are four sections; Mathematics, Physics, Chemistry and Biology in each part. Out of the four options given with each question, only one is correct.
PARTI (1 Mark Questions) 5. The parabola y = ax2 − 2 and y = 4 − bx2 intersect the
MATHEMATICS 1. How many positive integers less than 1000 are 6 times the sum of their digits? (a) 0
(b) 1
(c) 2
(d) 3
2. Divya inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of region between the two circles. Mansi did the same with a regular heptagon. The areas of two regions A and B respectively. Each polygon had a side length of 2. Which of the following is true? (a) 7A = 5B (c) A = B
(b) 5A = 7B (d) 25A = 49B
3. A box contains a collection of triangular and square tiles. There are 25 tiles in the box containing 84 edge total. The number of square tiles in the box are (a) 5
(b) 7
(c) 9
(d) 11
4. Define a function on the positive integers recursively by f (1) = 2, f (n ) = f (n − 1) + 2 if n is even, and f (n ) = f (n − 2) + 2 if n is odd and greater than one. Then, f (2019) is equal to
(a) 2019
(b) 2020
(c) 2021
(d) 2018
coordinate axes in exactly four points and these four points are the vertices of area 12, then a + b is equal to (a) 1/2
(b) 1
(c) 3/2
(d) 2
6. Let AB be a chord of circle with centre O. Let C be a point on the circle such that ∠ABC = 30° and O lies inside the ∆ABC. Let D be a point on AB such that ∠DCO = ∠OCB = 20°, then the measure of ∠CDO in degree is
(a) 110°
(b) 70°
(c) 80°
(d) 20°
7. Let a and b be natural numbers such that
2a − b, a − 2b and a + b are all distinct squares, the least possible value of b is (a) 21
(b) 22
(c) 24
(d) 25
8. The wealth of a person A equals the sum of that of B and C. If he distributes half of his wealth between B and C in the ratio 2 : 1, then the wealth of B equals the sum of that A and C. Then, the fraction of wealth that A should distribute between B and C in the ratio 1 : 2, so that the wealth of C equals the sum of that of A and B is (a)
1 2
(b)
2 3
(c)
3 4
(d) 1
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KVPY Practice Set 5 Stream : SA
9. For some positive integer K, the repeating baseK 7 representation of the (baseten) fraction is 51 0. 23K = 0232323 . K K , then the value of K is (a) 13
(b) 14
(c) 15
(c) melting point of solid (d) triple point of phase equilibrium
17. In the cyclic process, process A → B is isothermal. p A
(d) 16
10. The number 1, 2, 3, ..., 9 are randomly placed into the 9 square of a 3 × 3 grid. Each square gets one number and each of the numbers is used once. What is the probability that the sum of the numbers in each row and each column is odd?
(a)
1 21
(b)
1 14
(c)
5 63
(d)
2 21
C
B V
Correct V  T graph for the cycle is (b)
(a) V
11. A quadrilateral is inscribed in a circle of radius 200 2. Three sides of this quadrilateral have length 200, then length of the fourth side is (a) 200
(b) 200 2
(c) 400
V
B
C
(d) 500
A
C
A
B T
T
12. Let T be the smallest positive integer which, when divided by 11, 13, 15 leaves remainder in the sets { 7, 8, 9}, {(1,2,3}, {4, 5, 6} respectively. The sum of squares of the digit of T is (a) 50
(b) 81
(c) 89
(c) V
13. If roots of equation x − bx + c = 0 be two consecutive
A
(c) 2
(d) 1
the letters in the word MISSISSIPPI in which no two ‘S’ are together (b) 6 ⋅ 7 8 C4 (d) 6 ⋅ 8 ⋅7 C4
T
3 14 238 2 He , 7 N , 92 U 3 13 235 1 H , 6Cl , 92 U
train took 21 s to cross the platform which is 88 m long and that it took 9 s to pass him. Assuming that the train was moving with uniform speed. What is the length of the train in meters? (c) 66
Choose the correct statements given below. I. 2 He3 and 1 H3 are isotopes.
15. A man standing on a railway platform noticed that a
(b) 60
A
18. Consider the following nuclei:
14. How many different words can be formed by jumbling
(a) 55
C T
integers, then b − 4c equals
(a) 8 ⋅6 C4 ⋅7 C4 (c) 7 ⋅6 C4 ⋅8 C4
B
C
2
(b) 3
V
(d) 90
2
(a) −2
(d)
B
(d) 72
II.
235 92 U 13
III. 6Cl
and
238 92 U
are isobars.
14
and 7 N are isotones.
(a) All statements are correct (b) Both statements I and II are correct (c) Both statements II and III are correct (d) Only statement III is correct
19. Consider two identical copper spheres A and B. One
PHYSICS 16. p  T curve representing phase equilibrium is given by;
is placed over a thermally insulating plate,while the other hangs from an insulating thread.
p
A B
P1 T
The point P1 is (a) boiling point of liquid (b) condensation point of vapour
Equal amounts of heat are given to the two spheres and temperatures are recorded, then (a) TA = TB (c) TB > TA
(b) TB < TA (d) cannot be concluded
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KVPY Practice Set 5 Stream : SA
25. Given, A = Boltzmann constant, B = Planck’s constant
20. A boy throws a stone to hit a pole at some distance.
and C = speed of light. Then, quantity with dimensions of A4B−3C −2 is
Kinetic energy K of stone varies with horizontal displacement x as shown in figures given below. (b)
(a) K
K
x
x (c)
K
(d)
(a) universal gas constant (b) specific heat capacity (c) Stefan’s constant (d) heat energy
26. Considering air resistance, if t1 = time for a thrown ball in upward journey and t2 = time taken for downward journey, then
K
(a) t1 = t2 (c) t2 > t1 x
x
21. Ratio of time periods of small oscillations of the insulated spring and mass system (shown) before and after charging the mass is m
m
(a) equal to one (c) less than one
(b) greater than one (d) greater than or equal to one
22. The lights on a car are inadvertently left on. They dissipate 95 W. Fully charged 12 V car battery is rated 150 Ah. Time after which the car lights go OFF due to battery run down is (a) 12 h
(b) 24 h
(c) 18 h
(d) 36 h
23. In the arrangement shown below.
(b) t1 > t2 (d) 3 t2 = 2 t1
27. A car accelerates from rest at a constant rate α for sometime after which it deaccelerates at a constant rate β to come to rest. If total time is t, then maximum speed of car is α2 + β2 α2 − β2 α + β αβ t (d) t (a) t (c) t (b) α + β αβ αβ αβ
28. A lawn roller of mass 10 kg, radius 1 m is pulled horizontally by a handle attached to axle of the roller. F 0.4 m
Necessary minimum pull to raise roller above a step of 0.4 m is (a) 128 N (c) 213 N
(b) 134 N (d) 112 N
29. Geodesic is a (a) straight line (b) curve (c) circle (d) may be a straight line or curve
A m 2m B
30. In given combination of lenses, a parallel beam is
C
Accelerations of masses A and B just after cutting the string C are (a) 0, g
(b) g , g
g (c) , g 2
made incident from left as shown below. f= 1 m
f=0.25 m
(d) 2g , g
24. For streamlined flow of water, consider the following 0.75 m
statements. I. Two streamlines does not cross each other. II. Streamlines are straight.
Emerging light rays are as shown by (b)
(a)
(c)
(d)
III. Streamlined flow is more likely for fluids with low density and high viscosity. IV. Streamlined flow is more likely for liquids with high density and low viscosity. Which of the above statements are correct? (a) Statements I and III are correct (b) Statements II, III and IV are correct (c) Statements III and IV are correct (d) Statements I, III and IV are correct
CHEMISTRY 31. Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt. of Fe is 56) present in one molecule of haemoglobin is closest to (a) 1
(b) 6
(c) 4
(d) 2
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KVPY Practice Set 5 Stream : SA
32. Which of the following statements is incorrect? (a) Angular quantum number signifies the shape of the orbital (b) Energies of stationary states in hydrogen like atoms is inversely proportional to the square of the principle quantum number (c) Total number of nodes for 3sorbital is three (d) The radius of first orbit of He+ is half that of the first orbit of hydrogen atom
42. Sodium peroxide which is a yellow solid, when exposed to air becomes white due to the formation of (b) Na 2O (d) NaOH and Na 2CO3
(a) H 2 O2 (c) Na 2O and O3
43. The products formed when the following compound is treated with Br2 in the presence of FeBr3 are CH3
33. The solubility of saturated solution of calcium
fluoride is 2 × 10−4 mol L−1. Its solubility product is closest to
(a) 12 × 10−2 M 3 (c) 22 × 10−11 M 3
34. The brown ring complex compound is formulated as [Fe(H2O)5 (NO )] SO 4 . The oxidation state of iron is (a) 1
(b) 2
CH3
(b) 14 × 10−4 M 3 (d) 32 × 10−12 M 3
(c) 3
CH3
CH3 Br
(a)
and CH3
(d) 0
CH3
35. Which one of the following has the maximum dipole
Br
moment? (b) CH 4
(a) CO2
(d) NF3
(c) NH 3
CH3
CH3
Br
36. Which of the following is a chiral? (a) 1,1dibromo1chloropropane (b) 1, 1  dibromo 3chloropropane (c) 1, 3dibromo1chloropropane (d) 1, 3dibromo2chloropropane
Br
(b)
and CH3
CH3
CH3
compounds is I. phenol, III. pnitrophenol
II. ocresol IV. pchlorophenol
(a) I > II > III > IV (c) IV > III > II > I
(b) III > IV > I > II (d) III > II > IV > I
CH3
Br
37. The correct order of acidic character of the following
and
(c) CH3
CH3 Br
CH3
38. Which of the following is the correct order of size of
CH3
the given species? (a) I > I− > I+ (c) I > I+ > I−
(b) I+ > I− > I (d) I− > I > I+
(d)
and CH3
39. Which of the following is the correct representation of GayLussac’s law?
Br
CH3
Br
44. How many enantiomeric pairs are obtained by (a) p
monochlorination of 2, 3dimethyl butane ?
(b) p 1/V
(a) 4
of HCl. (a) − 184
(b) − 92
(c) 170
(d) − 88
T
40. A compound that gives a positive iodoform test is (a) pentanol (c) pentan2one
(b) pentan3one (d) pentanal
41. Which of the following compounds of xenon has pyramidal geometry? (a) XeOF4
(d) 1
H—Cl bond are 243, 435 and 431 kJ mol −1 respectively, then calculate the ∆ f H ° in kJ mol −1
(d) p 1/T
(c) 3
45. If bond enthalpies of Cl—Cl bond, H—H bond and
V
(c) p
(b) 2
(b) XeF2
(c) XeO3
(d) XeF4
BIOLOGY 46. Which of the following components provides sticky character to the bacterial cell? (a) Cell wall (c) Plasma membrane
(b) Nuclear membrane (d) Glycocalyx
223
KVPY Practice Set 5 Stream : SA 47. Life cycle of Ectocarpus and Fucus respectively are (a) haplontic, diplontic (b) diplontic, haplodiplontic (c) haplodiplontic, diplontic (d) haplodiplontic, haplontic
48. Which of the following are not polymeric? (a) Nucleic acid (c) Polysaccharides
flagellated cells called
the release of
(b) primary phloem (d) periderm
52. Which of the following options best represents enzyme composition of pancreatic juice? (a) Amylase, peptidase, trypsinogen, renin (b) Amylase, pepsin, trypsinogen, maltase (c) Peptidase, amylase, pepsin, renin (d) Lipase, amylase, trypsinogen, procarboxypeptidase
53. Which of the following are found in extreme saline conditions? (a) Archaebacteria (c) Cyanobacteria
(b) Fucus (d) Chlamydomonas
antagonist to gibberellins? (a) Zeatin (c) ABA
(b) Ethylene (d) IAA
57. The ornithine cycle removes two waste products from (b) atrial natriuretic factor (d) ADH
51. The vascular cambium normally gives rise to (a) phelloderm (c) secondary xylem
(a) Marchantia (c) Funaria
56. Which one of the following generally acts as an (b) oscula (d) mesenchymal cells
50. A decrease in blood pressure/volume will not cause (a) renin (c) aldosterone
(a) Chromosomes will not condense (b) Chromosomes will be fragmented (c) Chromosomes will not segregate (d) Recombination of chromosome arms will occur
55. Zygotic meiosis is a characteristic of
(b) Proteins (d) Lipids
49. In case of poriferans, the spongocoel is lined with (a) ostia (c) choanocytes
of animal cells. If APC is defective in a human cell, which of the following is expected to occur?
(b) Eubacteria (d) Mycobacteria
54. Anaphase Promoting Complex (APC) is a protein degradation machinery necessary for proper mitosis
the blood in liver. These products are (a) CO2 and urea (c) CO2 and ammonia
(b) ammonia and urea (d) ammonia and uric acid
58. Phellogen and phellem respectively denote (a) cork and cork cambium (b) cork cambium and cork (c) secondary cortex and cork (d) cork and secondary cortex
59. Which one of the following enzymes shows the greatest substrate specificity? (a) Lipase (c) Pepsin
(b) Nuclease (d) Sucrose
60. Albinism in humans is controlled by a recessive allele. How many copies of this allele will be found at one of the poles of a cell at telophaseI of meiosis in an albino person? (a) 23
(b) 4
(c) 2
(d) 1
PARTII (2 Marks Questions) MATHEMATICS
64. If ∆ABC is a right angle triangle with ∠ACB as its
61. A positive integer K is said to be good if there exists a partition of {1, 2, 3, K , 20} in to disjoint proper subsets such that the sum of the numbers in each subset of the partition is K. Then good number are there (a) 5
(b) 6
(c) 7
(d) 4
62. Let C1 and C 2 be circles defined by (x − 10)2 + y2 = 36 and (x + 15)2 + y2 = 81 respectively. The length of the shortest line segment PQ that is tangent C1 at P and to C 2 at Q is (a) 15
(b) 18
(c) 20
not both zero, and the repeating decimal 0. ab is expressed as a fraction in lowest terms. Then, the different denominators are possible, are (b) 4
(c) 6
(a)
1 2
(b)
2− 2 2
(c)
3− 3 3
(d)
5− 5 5
65. Let P (x) = (x − 1) (x − 2) (x − 3). For how many polynomials Q(x) does there exist a polynomial R(x) of degree 3 such that P (Q(x)) = P (x) ⋅ R(x)? (a) 22
(b) 24
(c) 27
(d) 32
(d) 24
63. Suppose that a and b are digits, not both nine and
(a) 3
right angle. The measure of ∠ABC = 60° and AB = 10. Let P be randomly chosen inside ABC, and extend BP to meet AC at D. Then, the probability that BD > 5 2 is
(d) 5
PHYSICS 66. Two stones are thrown up simultaneously from the
edge of a cliff 200 m high with initial speeds 15 ms −1 and 30 ms−1.
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KVPY Practice Set 5 Stream : SA
Correct graph of time variation of the relative position of the second stone with respect to first is (b)
(x2–x1) m
(x2–x1) m
(a)
8
8
(c)
This calorimeter is heated over a slow burner which provides heat at a constant rate. Temperature of calorimeter and its contents varies with time as shown below. T(°C)
t(s)
10
70. A calorimeter contains some ice and 10 kg water.
10
10
t(s)
0
(x2–x1) m
(x2–x1) m
(d)
8
10
–5
(a) 17 kg
t(s) 8
10
y (–22.5,0) (–30,0)
(b) 14 kg
(c) 10 kg
(d) 5 kg
t(s)
mirror by uv method, a student prepares following graph of u versus v graph.
(–45,0)
t(min)
60
Amount of ice initially present is nearly
67. In an experiment of finding focal length of a concave
u(cm)
50
x (0, –22.25) (0, –30) (0, –45)
CHEMISTRY 71. The heat liberated from the combustion of 0.5 g of carbon raised the temperature of 2000 g of water from 24°C to 26°C. The heat of combustion of carbon (per mole) is (a) −4 kcal (c) −62 kcal
(b) −8 kcal (d) −96 kcal
72. An organic compound of molecular formula C4H6.
A forms precipitates with ammoniacal silver nitrate and ammoniacal cuprous chloride. A is an isomer B, one mole of which reacts with one mole of Br2to form 1, 4dibromobut2ene. A and B are
(a) CH 3 CH 2 C ≡ CH and CH 2 == CHCH == CH 2 (b) CH 3 C ≡≡ CCH 3 and CH 3 CH == C == CH 2
CH2
v(cm)
(c)
Focal length of mirror is nearly (a) − 45 cm
(b) − 30 cm
(c) − 22.25 cm(d) − 15 cm
68. In given circuit, bulb that glows with maximum intensity is 2Ω
CH2 and
CH2
CH
CH2
CH CH
(d) CH3C
CCH3 and CH2
CH2 CH
4Ω
73. A gas bulb of 1 mL capacity contains 20 . × 1021
3Ω
molecules of nitrogen exerting a pressure of 757 . × 103 Nm −2. The root mean square speed of the gas molecules is
5Ω 6Ω
(a) 274 ms−1 (c) 690 ms−1
+ – 20 V
(a) 4Ω bulb (c) 3Ω bulb
the view point of molecular orbital theory?
of its displacement x is kx2 U (x) = 2 where, k = spring constant = 05 . Nm−1. If total energy of the particle is 1 J, then maximum amplitude of oscillation of particle is (b) 2 m
(c) 3 m
(b) 494 ms−1 (d) 988 ms−1
74. Which of the following statements is not correct from
(b) 2Ω bulb (d) 6Ω bulb
69. For a linear oscillator, potential energy as a function
(a) 1 m
CH2
C
(d) 1.5 m
(a) Be2 is not a stable molecule (b) He2 is not stable, but He+2 is expected to exist (c) Bond strength of N 2 is maximum amongst the homonuclear diatomic molecule belonging to the second period (d) The order of energies of molecular orbitals in N 2 molecule is σ 2s < σ * 2s < σ 2 pz ( π 2 px ≈ π 2 py ) < ( π * 2 px ≈ π * 2 py ) < σ * 2pz
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KVPY Practice Set 5 Stream : SA
The most likely explanation of the fact that the graph levels off at 180 Jm −2s−1 is that the system is
75. The atomicity of a molecule, M, if 10 g of it combine with 0.8 g of oxygen to form an oxide is closest to [specific heat of molecule, M is 0.033 cal deg −1 g −1 and molecular mass of molecule is 199.87 g] (a) 1 (b) 2 (c) 3 (d) 8
(a) light limited and carbon dioxide saturated (b) light limited and the temperature is below optimum (c) light saturated and carbon dioxide is unlimited (d) light saturated and the temperature is above optimum
BIOLOGY
78. Certain drug acts at synapses and affects the action of neurotransmitter substances. The table shows the effects of four different drugs.
76. The following statements describe the structure of certain protein molecule.
Drug Effect
(I) The molecule consists of two polypeptide chains which are folded around each other.
I. Inhibits the enzyme cholinesterase. II. Prevents the release of acetylcholine.
(II) In each chain the amino acids are held in a helix by hydrogen bonds.
III. Competes with acetylcholine at receptor sites.
Which orders of structure are described by these statements?
IV. Inhibits the enzyme which destroys noradrenaline.
Statement (I)
Which two drugs would prevent a skeletal muscle from responding to an electrical stimulus in the presynaptic neuron?
Statement (II)
(a) Primary (b) Secondary (c) Tertiary (d) Quaternary
Tertiary Tertiary Secondary Secondary
(a) I and II (c) II and III
(b) I and IV (d) II and IV
79. The diagram shows some chromosomes at late
77. The graph shows the relationship between oxygen
prophase of mitosis.
production in photosynthesis and light intensity for a unicellular green organism in 0.02% sodium hydrogencarbonate solution Oxygen production/mm3h–1 7 6 5
How many chromosomes would be present in one nucleus at telophaseII of meiosis?
4
(a) 6 (c) 18
3 2
80. Pyrimidine bases contain four carbon atoms and
1 0 –1
(b) 12 (d) 24
purine bases contain 5. How many carbon atoms are there in a nucleotide containing cytosine?
20 40 60 80 100 120 140 160 180 200 Light intensity/Jm–2S–1
(a) 8
–2
(b) 9
(c) 10
(d) 11
Answers PARTI 1 11 21 31 41 51
(b) (d) (a) (c) (c) (c)
2 12 22 32 42 52
(c)
62 72
(c)
(b) (c) (c) (d) (d)
3 13 23 33 43 53
(c)
63 73
(d)
(d) (d) (d) (c) (a)
4 14 24 34 44 54
(b)
64 74
(c)
(c) (a) (a) (d) (c)
5 15 25 35 45 55
(c)
65 75
(a)
(c) (c) (c) (b) (d)
6 16 26 36 46 56
(c)
66 76
(d)
(d) (c) (c) (d) (c)
7 17 27 37 47 57
(a)
67 77
(d)
(a) (a) (b) (c) (b)
8 18 28 38 48 58
(d)
68 78
(a)
(d) (b) (d) (d) (b)
9 19 29 39 49 59
(d)
69 79
(b)
(c) (d) (d) (c) (d)
10 20 30 40 50 60
(b)
70 80
(a)
(c) (b) (c) (d) (c)
PARTII 61 71
(b) (d)
(a)
(b)
(d)
(a)
(d)
(d)
(c)
(a)
(b)
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KVPY Practice Set 5 Stream : SA
Solutions abc = 100a + 10b + c where a , b, c ∈ {0, 1, 2, 3, ..., 9} and a + b + c> 0 The sum of digits of this number is (a + b + c). Given, 100a + 10b + c = 6(a + b + c) ∴ 94a + 4b − 5c = 0 Clearly, a > 0. No solution ∴ a = 0 then 4b = 5c This is possible only b = 5 and c = 4 is 54. ∴Number Hence, only one number i.e., 54.
=x and the number of square tiles = y A triangle has three edges and square has four edges. ...(i) ∴ x + y = 25 and ...(ii) 3x + 4 y = 84 On solving Eqs. (i) and (ii), we get x = 16, y = 9 Hence, number of square tiles in box is 9.
2. (c) In ∆OPB, cos
π r π = ⇒ r = R cos 5 R 5
4. (b) We have, f (1) = 2 and f (n ) = f (n − 2) + 2, n is odd ∴ f (3) = f (1) + 2 = 2 + 2 = 4 f (5) = f (3) + 2 = 4 + 2 = 6 ∴ Similarly, f (2019) = 2020
5. (c) We have, Equation of parabola
2π — 5
and
y = ax2 − 2
…(i)
y = 4 − bx
…(ii)
O
2
Y
r
C (0, 4)
R P B
Area of region = π (R 2 − r 2 ) π = πR 2 1 − cos2 5 π 2 π 1 − cos 5 5 Q sin π = 1 5 R 2 π 2 π = π cosec − cot = π 5 5
O
X′ (x2, 0) D
= π cosec2
A (0, –2) Y′
On solving Eqs. (i) and (ii), we get 6 x intercept are ± a+ b i.e. coordinate of points 6 A (0, − 2), B , 0 , C (0, 4) a + b
Similarly in heptagon,
and D = −
π/7
r1
X B (x1, 0)
R1
π Area of region = π cosec 7 = π cosec2 2
= π ∴ Both have same area. ∴ A=B
6 a+ b
1 AC × BD 2 6 6 1 =2 12 = 6 × 2 ⇒ a+ b 2 a+ b 6 =4 a+ b 6 3 a+ b= = 4 2
Area of kite ABCD =
1 − cos2 π 7 π π − cos2 7 7
⇒ ⇒ ⇒
6. (c) Given, ∠ABC = 30° ∴
∠AOC = 2 ∠ABC = 60°
C
20°
3. (c) Let the number of triangular tiles
20°
1. (b) Number less than 1000 can write
O A
D
30° B
Now, OA = OC ∆OAC is an equilateral ⇒ ∠CAO = ∠ACO = 60° ⇒ ∠ACD = 60° − 20° = 40° OC = OB ∴ ∠OBC = ∠OCB = 20° ⇒ ∠OBA = 10° = ∠OAB ⇒ ∠DAC = 70° In ∆ACD, ∠CDA = 70° ⇒ ∠CDA = ∠CAD = 70° ⇒ CD = CA = CO In ∆CDO , CD = CO and ∠DCO = 20° 180° − 20° ⇒ ∠CDO = = 80° 2 7. (a) We have, 2a − b, a − 2b and a + b are squares. ...(i) ∴Let 2a − b = x2 ...(ii) a − 2b = y2 ...(iii) and a + b = z2 where x, y, z ∈ N From Eqs. (ii) and (iii), we get 2a − b = y2 + z 2 ...(iv) ∴ x2 = y2 + z 2 From Eqs. (i) and (iii), we get 3a = x2 + z 2 2 2 x + z is multiple of 3 ⇒ x and z is also multiple of 3. From Eqs. (ii) and (iii), we get ...(v) 3b = z 2 − y2 2 2 z − y is a multiple of 3 ⇒ y and z is also multiple of 3. Let x = 3x1 , y = 3 y1 , z = 3z1 ⇒ x12 = y12 + z12 Let us assume every two of x, y, z are coprime. ⇒ x1 , y1 , z1 are pythagorean triplet. ⇒ Out of y1 and z1 , one even ≥ 4 and other odd ≥ 3. From Eq. (v), we get b = 3(z12 − y12 ) = 3(z1 + y1 ) (z1 − y1 ) ⇒ min b = 3(4 + 3) (4 − 3) = 21
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KVPY Practice Set 5 Stream : SA
4ax = 4x ⇒ a = 1
0. 23 K as follows 0. 23 K = 2 ⋅ K −1 + 3 ⋅ K −2 + 2 ⋅ K −3 + 3 ⋅ K −4 + K −1 −3 −5 = 2(K + K + K + ...) + 3(K −2 + K −4 + K −6 + K) 2 1 1 = 1 + 2 + 4 + K K K K 3 1 1 + 2 1 + 2 + 4 + K K K K 2 3 1 2K + 3 = = + 2 K K 1 − 1 K 2 − 1 2 K 7 Given, 0. 23K = 51 2K + 3 7 ∴ = 2 K − 1 51 ⇒ 7K 2 − 102K − 160 = 0 ⇒ 7K 2 − 112K + 10K − 160 = 0 ⇒ (7K + 10) (K − 16) = 0 −10 ⇒ K = 16, K ≠ 7
10. (b) Sum odd only be formed (even, even, odd) or (odd, odd, odd). So can focus on placing evens, we need to have
AB = BC = CD = 200 OA = OB = OC = OD = 200 2 ∠AOB = ∠BOC = ∠COD B
200
C
A
0
⇒
9. (d) We can expand the fraction
11. (d) Given,
20
B=A+C x x x x y+ = + z+ ⇒ y−z= ∴ 3 2 6 3 Let ‘a’ be the fraction that A should distribute and the ratio of distribution is 1 : 2. 1 2 < 3 3 Now, A = (1 − a ) x ax B= y+ 3 2ax C=z+ 3 2ax ax z+ = y+ + (1 − a ) x ∴ 3 3 ⇒ 3z + ax = 3 y + 3x − 3ax ⇒ 4ax = 3( y − z ) + 3x x x 4ax = 3 + 3x Q y − z = ⇒ 3 3 Now,
each even be with another even in each row or column. There are 9 ways to this. They are 5! ways to permute odd and 4! ways to permute even number. 5! × 4! × 9 1 ∴Required probability = = 9! 14
0
x, y and z respectively. Given, A=B+C ∴ x= y+ z A distributes half of his wealth to B and C in the ratio 2 : 1. 2 x 1 x Wealth of B = y + and C = z + 3 2 3 2
20
8. (d) Let the wealth of A, B and C are
E
F
O
D
√2 20 0
In ∆OAB and ∆ABE, ∠BAE = ∠AOB ∠ABO = ∠ABE ∴ ∆OAB ~ ∆ABE OA AB OB = = AB BE AE [QOA = OB ] AB = AE Similarly in ∆OCD and ∆DFE, CD = DF OB ∴ OE = 100 2 = 2 BC and EF = = 100 2 ∴ AD = AE + EF + FD = 200 + 100 + 200 = 500
14. (c) We have, MISSISSIPPI Other than S, seven letters, M, I, I, I,P, P, I can be arranged in 7! = 7× 5× 3 4! 2! Now, 4 S can be placed in 8 spaces, i.e. 8 C4 . ∴Total number of arrangement = 7 ⋅ 5 ⋅ 3 ⋅8 C4 = 7 × 15 × 8C4 = 7 × 6C4 × 8C4
15. (c) Let the length of train is x m. Length of train and platform = (x + 88) m Time taken by train to cross the platform = 21s x + 88 ...(i) Speed = ∴ 21 Time taken by train to cross the man = 9 s x Speed = ...(ii) ∴ 9 From Eqs. (i) and (ii), we get x + 88 x = ⇒12x = 88 × 9 21 9 88 × 9 x= = 66 m ⇒ 12 16. (d) Point P1 is called triple point, where fusion curve vaporisation curve and sublimation curve meets. 17. (a) Process AB is isothermal expansion, so V B A
12. (b) T = {4, 5, 6} (mod 15)
T
or T = {19, 20, 21}, {34, 35, 36}, {49, 50, 51}, {64, 65, 66}
Process BC is isobaric, so
{79, 80, 81}, {94, 95, 96}, {109, 110, 111}, {124, 125, 126}
B
{139, 140, 141}, {154, 155, 156}, {169, 170, 171}, {184, 185, 186} (mod15) Now, by direct checking we get smallest T = 184 ∴ Required sum = 1 + 8 + 4 = 81 2
2
2
13. (d) We have, x2 − bx + c = 0 Let α , β are the roots of the equations. ∴ α + β = b, αβ = c Given, α −β =1 ∴ (α + β )2 − (α − β )2 = 4αβ b2 − 1 = 4c 2 ⇒ b − 4c = 1
V
C T
18. (d) Number of neutrons, N =A−Z For 6 Cl13 , N = 13 − 6 = 7 and 7 N14 , N = 14 − 7 = 7 So, they contains same number of neutrons.
19. (c) Part of heat given to A is used up in doing work (against) gravitational force. So, temperature of B will be slightly higher. 20. (c) Kinetic energy of stone decreases then increases. It is never zero in entire flight of stone.
228
KVPY Practice Set 5 Stream : SA
21. (a) In absence of charge, time period m T1 = 2 π 2k In presence of charge, time period m T2 = 2 π 2k
If surface over which points are located is a plane, then Geodesic is a straight line. On the surface of a planet, Geodesic is a great circle joining two points.
30. (b) Image of first lens is formed at focal point of second lens. 0.25 m
So, ratio of time period is equal to one.
22. (c) Total output energy of battery = V (It ) = 12 V × 150 Ah = 12 × 150 × 3600 J = 6.48 × 106 J Energy consumed by car lights in time t = 95t Equating both values, we get t=
6.48 × 10 ≈ 18. 9 h 95 6
23. (d) When string C is cut, spring snaps back to regain its unstretched length. So, spring force on m remains same. Hence, accelerations of m and 2m are 2g and g, respectively. 24. (a) Streamlines may be straight or curved. In fluids, with low viscosity streamlined flow occurs only at low flow speed.
25. (c) [A 4 B −3 C 2 ] = [ML2 T −2K −1 ]4 ⋅ [ML2 T −1 ]−3 ⋅ [LT −1 ]−2 = [M4 − 3 L8 − 6 − 2 T −8 + 3
+ 2
K −4 ]
= [ML0 T −3 K −4 ] = Stefan’s constant.
26. (c) Time for downward journey is higher as ball can be thrown with any velocity but its downward velocity is always less than or equal to terminal velocity. 27. (a) Let car accelerates for time t1 and then it deaccelerates for time t2. Then, αt1 − βt2 = 0 and t = t1 + t2 αβ Maximum speed, v = αt1 = t α + β
28. (b) Taking moments about point of contact O.
So, emerging rays are parallel.
31. (c) 100 g of haemoglobin contains = 0.33 g Fe 67200 g of haemoglobin contains ∴ 0.33 × 67200 g Fe = 100 = 221.76 g Fe ∴ Number of Fe atom present in one 221.76 molecule of haemoglobin = ≈4 56
1m
O
signifies the shape of the orbital. e.g. When l = 0, the shape of orbital is spherical. When l = 1, the shape of orbital is dumbbell. Thus, statement (a) is correct. (b) According to Bohr’s theory For hydrogen like atom Z2 En = − 13.6 2 eV n Thus, statement (b) is correct. (c) Total number of nodes in orbital = n − l−1 For 3 sorbital, n = 3, l = 0 Number of nodes = 3 − 0 − 1= 2 Thus, statement (c) is incorrect. (d) According to Bohr’s radius a n2 r= 0 Z a0 (1)2 a (1)2 rHe + = , rH o 2 1 Thus, statement (d) is correct. For the reaction, CaF2 Ca2+ (aq) + 2F − (aq)

0.8
F × 0.6 = 100 × 0.8 100 × 0.8 − 134 N F= = 133.3 N ~ ⇒ 0.6
29. (d) Shortest distance between two points is called Geodesic.
H
∴
S
µnet=0
34. (a) Let, the oxidation state of Fe in [Fe(H 2 O)5 (NO)] SO4 be x . ∴ x + 5 (0) + 1 = + 2 ⇒ x=+1
H
H µnet=0
H
N
F
H
N
H
F
µ net
µ net
= 4.90 × 10−30 cm
F
= 0.80 × 10−30 cm
36. (c) A molecule is set to be chiral, if all the four groups attached to central carbon atom are different.
Br 3
2
1
(a) H3C—CH2—C—Cl Br 1, 1dibromo1chloropropane It is an achiral molecule.
Br 3
2
1
(b) Cl—CH2—CH2—CH Br 1, 1dibromo3 chloropropane It is an achiral carbon.
Br 3
2
1
* BrCH2—CH2—CH
Cl 1, 3dibromo1chloropropane It is a chiral molecule with chiral carbon position at 1(*).
Cl Br
2S
Ksp = [Ca2+ ] [F − ]2 = (S) (2S)2 = 4S 3 Ksp = 4 (2 × 10−4 )3 = 32 × 10−12 M3
C
H
O==C==O
32. (c) (a) Angular quantum number, l
33. (d) Let the solubility of CaF2 be S. 0.6
35. (c) CO2 and CH 4 have zero dipole moment as these are symmetrical in nature. Between NH3 and NF3 , NH3 has greater dipole moment though in both NH3 and NF3 , N possesses one lone pair of electrons. This is because in case of NH3 , the net N—H bond dipole is in the same direction as the direction of dipole of lone pair but in case of NF3 , the direction of net bond dipole moment of three N—F is opposite to that of the dipole moment of the lone pair which cancel out the resultant dipole.
(d)
3 2 1 H2C—C—CH2
Br Cl 1, 3dibromo2chloropropane It is an achiral molecule.
229
KVPY Practice Set 5 Stream : SA 37. (b) When an electron withdrawing group, like NO2, Cl is attached to the phenol ring, it stabilises the negative charge on the oxygen of the phenoxide ion. Due to this reason, acidic character of phenol increases. Between compound III and IV, III is more acidic as NO2 is more stronger EWG than Cl. But when an electron donating group, like CH3 is attached to the phenol ring, it destabilises the ring and hence acidic character of phenol decreases. Thus, the correct order of acidic character is
OH
OH
OH
OH CH3
>
>
As it contains C CH3 group. O Thus, it will show positive iodoform test. (d) CH 3 CH 2 CH 2 CH 2 CHO (Pentanal)
It gives negative iodoform test. 41. (c) The structures of given species are shown below:
O F
F F
Xe
F
Xe
F
F
XeOF4 Square pyramidal (sp3d2)
Cl
F
III
IV
Xe
Xe I
II
39. (d) According to Gay Lussac’s law, at constant volume, the pressure of given mass of the gas is directly proportional to its absolute temperature, i.e. p ∝ T or p = kT Thus, the correct representation is given in option (d).
42. (d) Sodium peroxide which is a yellow solid, reacts with moisture and CO2 of air (when exposed to air) and becomes white due to the formation of NaOH and Na2 CO3 . 2Na2 O2 + H 2 O → 4NaOH + O2 2NaOH + CO2 → Na2 CO3 + H 2 O
43. (c) As CH 3 group is orthopara directing, so the major products will be formed at o and pposition only. CH3 CH3 Br2/FeBr3
CH3
40. (c) Positive iodoform test are given by those carbonyl compounds which contain CH 3 C group. O (a) CH 3 CH 2 CH 2 CH 2 CH 2 OH (Pentanol)
It gives negative iodoform test. (b) H 3 C CH 2 C CH 2 CH 3 O (Pentan3one)
It does not show iodoform test. (c) H 3 C CH 2 CH 2 C CH 3 O (Pentan2one)
F
XeF4 Square planar (sp3d2)
XeO3 Pyramidal (sp3)
p T
F
O
38. (d) Anion is formed by the gain of electron to the neutral atom and cation is formed after the loss of electron from the neutral atom. Hence, cation has smaller size due to increased nuclear charge whereas anion has bigger size than its neutral atom. Thus, the correct order of size is I− > I > I+ .
O
Br
Electrophilic substitution
+ CH3
Br2/FeBr3
CH3
CH3 Br
CH3 Br
CH3 Not possible due to steric Hindrance Thus, the correct option is (c).
44. (d) CH 3 CH CH CH 3 + Cl 2 CH3
∆H = [(BE)H − H + (BE)ClCl ] − 2 [(BE)HCl ] = [435 + 243] − 2 [431] = − 184 kJ mol −1 The moles of HCl( g ) are formed from its element, hence − 184 ∆ f H °( HCl ) = = − 92 kJ mol −1 . 2 mucilage layer of the cell envelope. It gives sticky character to the bacterial cell.
>
O
45. (b) H2 ( g ) + Cl 2 ( g ) —→ 2HCl( g )
46. (d) Glycocalyx is the outermost
XeF2 L inear (sp3d2)
F NO2
Due to the presence of chiral centre (*), it shows the optical activity and its mirror image are nonsuperimposable. Hence, it shows one enantiomeric pair.
CH3
2, 3dimethyl butane
CH3 → CH 3 CH C H CH 2 Cl * CH3
1− chloro2, 3 dimethyl butane
47. (c) Ectocarpus and Fucus respectively show haplodiplontic and diplontic life cycle. In Ectocarpus, sporic meiosis occurs and haploid biflagellate meiozoospores are formed. They germinate to produce gametophytic thalli. The gametophytes liberate gametes which fuse to form diploid zygote which gives rise to a diploid plant. In Fucus, there is a single somatic phase. It is diploid and produces haploid gametes. They fuse during fertilisation to give rise to diploid individual. 48. (d) Among the given options, except lipids all are polymers. These are formed by the polymerisation of monomers. The basic unit of lipid are fatty acids and glycerol molecules that do not form repetitive chains. Instead they form triglycerides from 3 fatty acids and one glycerol molecules. Protein monomers are amino acids and they bond together in repetitive chains just as carbohydrate monomers are monosaccharides. 49. (c) The body wall of a common sponge consists of three layers i.e. pinacoderm, choanoderm and mesophyll layers. Choanoderm is inner cellular layer which consists of highly specialised flagellated cells called choanocytes. The beating of their flagella creates water current.
50. (d) A decrease in blood pressure/volume stimulates the hypothalamus to release ADH (Antidiuretic Hormone) as well as JGA cells to release renin. Renin by renin angiotensin mechanism activates the adrenal cortex to release aldosterone. ANF (Atrial Natriuretic factor) is produced by atria of heart during increased blood pressure. It can cause vasodilation and thereby decrease the blood pressure. Therefore, option (d) is correct.
230 51. (c) Vascular cambium located between xylem and phloem in the stems and roots of vascular plants. It produces secondary xylem towards the pith and secondary phloem towards the bark.
52. (d) Pancreas consists of exocrine and endocrine parts. Exocrine part secretes alkaline pancreatic juice. This juice contains trypsinogen, chymotrypsinogen, procarboxypeptidase, lipase, amylase, elastase.
53. (a) Archaebacteria are the most primitive form of bacteria. These live in diverse habitat, e.g. extreme hot temperature, saline condition, variable pH, etc. Saline bacteria are called halophiles (e.g. Halobacterium, Halococcus).
54. (c) If anaphase promoting complex is defective in a human cell, the chromosome will not segregate during anaphase of mitosis. APC triggers the transition from metaphase to anaphase by tagging specific proteins for degradation.
55. (d) Zygotic meiosis is represented in the haplontic life cycle of many algae including Chlamydomonas. In such a life cycle, all cells are haploid except zygote. This is because meiosis occurs in the zygote itself resulting into four haploid cells that give rise to haploid plants.
56. (c) Gibberellins and ABA are
KVPY Practice Set 5 Stream : SA pair of sister chromatids at this stage each chromosome and hence each end would have two copies of the allele.
61. (b) Let us partition in to n parts and each part has sum = K, then nK = 1 + 2 + 3 + K + 20 ⇒ nK = 210 ∴K divides 210. Also, K must be ≥ 20 . Now, 210 = 2 × 3 × 5 × 7 So, proper divisor of 210 are {1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210} ⇒ K can be 21, 30, 35, 42, 70,105 For K = 21, we have (1, 20) (2, 19) ... (10, 11) ⇒ 21 is a good number. For K = 42, join twotwo pairs For K = 105, join fivefive pairs ⇒ 42 and 105 are also good numbers For K = 30, we have {20, 10}, {19, 11}, {18, 12}, {17, 13}, {16, 14}, {15, 9, 6}, {1, 2, 3, 4, 5, 7, 8} ⇒ K = 30 is also good number Similarly, 35 and 70 also good numbers. ∴There are total 6 good numbers.
62. (c) We have, C1 : (x − 10)2 + y2 = 36 C2 : (x + 15)2 + y2 = 81 Centre of C1 = (10, 0) and radius = 6 Centre of C2 = (−15, 0) and radius = 9
antagonistic to each other. Gibberellins promote seed germination whereas ABA promotes seed dormancy.
Q 9 9
57. (b) Ornithine cycle removes both
6
A(–15,0)
B (10,0)
ammonia and urea from the blood. It converts ammonia into urea (in liver) and transports it to kidneys by the blood. Hence, it plays a key role in detoxification of our blood. This cycle occurs in the liver.
The length of smallest line segment PQ is the indirect common tangent of circle
58. (b) In the dicot stem, the cortical
∴
PQ =
AB 2 − (r1 + r2 )2
⇒
PQ =
(25)2 − (9 + 6)2
cells get differentiated to give rise to another meristematic tissue, which is called cork cambium or phellogen. On the other side, it forms phellem (cork) and in the inner region it forms secondary cortical cells (phelloderm).
59. (d) Sucrose shows the most substrate specificity since it hydrolyses only the disaccharide, sucrose.
60. (c) Since the allele is recessive, both homologous chromosomes in a somatic cell of an albino person would have the allele. After meiosisI, each end would have a homologous chromosome with the allele. As the chromosome is existing as a
P
[Q AB = ⇒ ⇒
PQ = PQ =
6
(10 + 15)2 − 02 , AB = 25] 625 − 225 400 = 20
63. (d) The repeating decimal 0.ab is equal to ...(i) x = 0. abababab ...(ii) 100x = ab ⋅ abababab On subtracting Eq. (i) from Eq. (ii), we get 99x = ab ab 10a + b ⇒ x= = 99 99
When expressed in lowest term, the denominator of this fraction will always be a divisor of 99 = 3 ⋅ 3 ⋅ 11 This gives us the possibilities {1, 3, 9, 11, 33, 99}. As a and b both are not both 9 and not both zero the denominator 1 cannot be possible. ∴ Possible denominators are {3, 9, 11, 33, 99}.
64. (c) In ∆ABC, ∠C = 90°, AB = 10, ∠B = 60° B 10
P
C
D
A
In ∆ABC, BC = AB cos B = 10 × cos 60° = 5 AC = AB sin B = 10 × sin 60° = 5 3 Choose a P ′ and get a corresponding D′ such that BD′ = 5 2 BD > 5 2 >
BC 2 + CD 2 >
25 + CD 2
⇒ CD > 5 Thus, the point P may only lie in the ∆ABD′ Area of ∆ABD ′ Required probability = Area of ∆ABC The ratio of AD′ to AC because the triangle have identical altitudes. AD ′ AC − CD ′ So, ratio = = AC AC 3−1 CD ′ 5 = 1− = 1− = AC 5 3 3 ∴Required probability ( 3 − 1) × 3 3 − 3 = = 3× 3 3
65. (a) We have, P (x) = (x − 1) (x − 2) (x − 3) and P[Q (x)] = P (x) ⋅ R (x) ∴ P[Q (x)] = [Q (x) − 1] [Q (x) − 2] [Q (x) − 3] = (x − 1) (x − 2) (x − 3) R (x) Since, degree of P (x) = 3 and degree of R (x) = 3 ∴Degree of [P (x) ⋅ R (x)] = 6 Thus, degree of P (Q (x)) = 6, so degree Q (x ) = 2 P (Q (1)) = (Q (1) − 1) (Q (1) − 2) (Q (1) − 3) = 0 P (Q (2)) = (Q (2) − 1) (Q (2) − 2) (Q (2) − 3) = 0 P (Q (3)) = (Q (3) − 1) (Q (3) − 2) (Q (3) − 3) = 0
Hence, we conclude Q (1), Q (2) and Q(3) must each be 1, 2, 3. Since, a quadratic is uniquely determined by the three points.
231
KVPY Practice Set 5 Stream : SA
66. (d) For first stone, 1 2 at 2 = 15t − 5t 2 [Q a = g = − 10 ms−2] Now, x1 = − 200 m ⇒ 15t − 5t 2 = − 200 2 ⇒ 5t − 15 t − 200 = 0 ⇒ 5t 2 − 40t + 25t − 200 = 0 ⇒ 5t (t − 8) + 25 (t − 8) = 0 ⇒ t = 8 s or t = − 5 s (Q t = −5 s not acceptable) ∴Time for which first stone remains in air = 8 s. So, graph is x2 − x1 = 15t (For t = 0 to t = 8 s) and x2 − x1 = 200 + 30t − 5t 2 (For t > 8 s to t = 10 s) x1 = ut +
69. (b) At extreme positions, total energy is potential energy. kx2 0.5x2 So, U (x ) = ⇒1 = 2 2 ⇒ x2 = 4 ⇒ x = ± 2m So, particle turns back after reading 2 m mark. ∴Amplitude of oscillation of particle is 2 m. 70. (a) Let m = mass of ice initially present. Heat absorbed from t = 0 to t = 50 min is Q1 = miceL = m(kg) 333 (J/g) = m (3.33 × 105 ) J and heat absorbed from t = 50 min to t = 60 min, Q2 = ms∆T = (10 + m) (4186 . × 103 ) (10 − 0) = (10 + m) (4186 . × 103 ) (10) Given that, rate of heat supply is constant. Q1 Q So, = 2 ∆t1 ∆ t2
distance, image formed is also at 2f . So, from graph, 2f = − 30 cm ⇒ f = − 15 cm
m (3.33 × 105 ) (10 + m) (4186 . × 103 ) (10) = 50 10 m(33.33) = (10 + m)(4186 . ) ⇒ 5 ⇒ 33.33 m = 209.3 + 20.93 m 12.4 m = 209.3 ⇒ ⇒ m = 16.87 kg
68. (a) Circuit can be redrawn as
71. (d) Amount of heat liberated by
67. (d) When object is placed at 2f
2Ω A
0.05 g of C
4Ω 3Ω
6Ω
C
B
5Ω
+ – 20 V
As RAB and RAC are nearly 1 Ω and 2Ω respectively, a larger potential drop occurs across BC. As AB and BC are in
= mC∆T = 2000 × 1 (26 − 24) = 4000 cal = 4 kcal 4 kcal × 1 Calorific value = = 8 kcal per g 0.5 ∴ Heat of combustion = − 8 × 12 = − 96 kcal per mol
72. (a) As the organic compound forms white precipitate with AgNO3 /NH4 OH and red precipitate with Cu 2Cl 2/NH4 OH, it must be a terminal alkyne.
Thus, terminal alkyne (A) with molecular formula C4 H 6 is CH 3 CH 2 C ≡≡ CH. Thus, option (a) is correct.
CH2— CH—CH— CH2 B (Isomer of A) –
+
CH2 — CH — CH— CH2 Br2
CH2 — CH — CH— CH2
—
series and current remains same in both, so IRAB < IRBC . Now, for a parallel combination, V2 1 or P ∝ . P= R R Hence, a larger potential drop occurs across 4Ω bulb. Hence, 4Ω bulb glows brightest in given circuit.
—
There can be 3 × 3 × 3 = 27 different quadratic. However, we have included Q (x) which are not quadratic. They are line. Then, Q (1) = Q (2) = Q (3) = 1 ⇒ Q (x ) = 1 Q (1) = Q (2) = Q (3) = 2 ⇒ Q (x ) = 2 Q (1) = Q (2) = Q (3) = 3 ⇒ Q (x ) = 3 Q (1) = 1, Q (2) = 2, Q (3) = 3 ⇒ Q (x ) = x Q (1) = 3, Q (2) = 2, Q (3) = 1 ⇒ Q (x ) = 4 − x So, these linear function are not included ∴ Total number of polynomials = 27 − 5 = 22
Br
Br
1, 4 dibromobut2ene
73. (b) Number of moles of the gas =
2.0 × 1021 6.023 × 1023
mol
= 3.32 × 10−3 mol From ideal gas equation pV = nRT 7.57 × 103 × 10−3 pV = T = nR 3.32 × 10−3 × 8.314 = 274.25 K Root mean square speed, 3RT vrms = M 3 × 8.314 × 274.25 = 28 × 10−3 = 494.26 ms−1
74. (d) (a) Electronic configuration of Br2 = σ1s2 σ * 1s2 σ 2s2 σ * 2s2 4− 4 B.O = =0 2 Thus, it does not exist. Hence, statement (a) is correct. 2− 2 (b) He 2 = σ1s2 σ * 1s2 B.O = =0 2 Hence, it will not exist. 2−1 = 0.5 He +2 = σ1s2σ * 1s2, B.O = 2 Since, the bond order is not zero, this molecule is expected to exist. (c) N 2 = σ1s2σ * 1s2 σ 2s2 σ * 2s2 π 2 px2 ≈ π 2 py2 σ 2 pz2 10 − 4 B.O = =3 2 Thus, it has the maximum bond strength among the other diatomic molecule belonging to the second period. (d) The correct electronic configuration of N 2 is σ1s2 σ * 1s2σ 2s2 σ * 2s2 π 2 px2 ≈ π 2 py2 σ 2 pz2 Thus, statement (d) is incorrect.
232 75. (a) Equivalent mass of M mass of metal = ×8 mass of oxygen 10 = × 8 = 100 0.8 Approximate atomic mass 6.4 6.4 = = = 193.3 g specific heat 0.033 193.93 Valency of M = = 1.98 ≈ 2 100 So, accurate atomic mass = equivalent mass × valency = 100 × 2 = 200 g molar mass 199.87 Atomicity = = =1 atomic mass 200
76. (d) The folding of two or more polypeptide chains constitutes the
KVPY Practice Set 5 Stream : SA quaternary structure. Tertiary structure of protein is a single polypeptide chain folded and twisted. Primary structure of protein is amino acids joined end to end with each other. Secondary structures are folded into αhelix and βpleated sheets.
79. (a) There are 12 chromosomes present. In a mitotic division, the number of chromosomes remains the same, i.e. 2n → 2n. In a meiotic division, the number is halved, i.e., 2n → n so, there should only be 6 chromosomes in the nucleus.
77. (d) The amount of light given is
80. (b) Phosphate group, ribose or deoxyribose group and cytosine group form the nucleotide. Phosphate (H 3 PO4 ) Ribose (C 5 H10O5 ) Deoxyribose (C 5 H10O4 ) Cytosine has 4 carbon atoms, as it is a pyrimidine. Therefore, the nucleotide should have 5C + 4C = 9C
saturated, not limited. Therefore the answer is (d). The concentration of carbon dioxide is in short supply, hence limiting the rate of photosynthesis.
78. (c) The drugs will only prevent the response of the muscle to an electrical stimulus if it prevents the release of acetylcholine and so inhibits the increase in membrane permeability to sodium ions, and if the drug competes with acetylcholine at the receptor sites.